The nurse's push must equate to the frictional force for the cart to move at a constant velocity. Since the push is at an angle, its horizontal component, which opposes friction, is calculated and set equal to the friction. Solving for the force exerted by the nurse yields the answer.
Explanation:This is a physics problem related to force, friction, and motion. To keep the cart at a constant velocity, the net force exerted on it must be zero, as per Newton's second law. This means that the frictional force, which opposes the direction of motion, should be equal to the applied force, which is the nurse's push.
The nurse's push is exerted at an angle, so we have to decompose it into horizontal and vertical components. The horizontal component is the one that actively works against the friction and it can be calculated using the formula: F_Horizontal = F_Nurse cos θ, where 'F_Nurse' is the force applied by the nurse and 'θ' is the angle of the push. Since this component must equal the frictional force, we can rearrange the equation and solve for 'F_Nurse': F_Nurse = Friction / cos θ. Plugging in the known values for friction (65 N) and angle (36 degrees), we can find the magnitude of force the nurse must exert to move the cart at a constant velocity in newtons.
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To move the cart at a constant velocity, the nurse must exert a force of approximately 79.3 Newtons at a downward angle of 35 degrees below the horizontal. The force of friction and the horizontal component of the nurse's force must be equal for the cart to move at a constant velocity.
Explanation:To calculate the force the nurse must exert to move the cart at a constant velocity, we need to consider the forces acting on the cart. In this case, the force of friction and the force exerted by the nurse are the main forces. The force exerted by the nurse can be broken down into horizontal and vertical components. The horizontal component of the force is responsible for overcoming the force of friction. Since the cart is moving at a constant velocity, the net force in the horizontal direction is zero. To find the magnitude of the force the nurse must exert, we can use trigonometry to find the horizontal component of the force.
The horizontal component of the force can be found using the equation:
F_h = F_n * cos(theta)
Where F_h is the horizontal component of the force, F_n is the total force exerted by the nurse, and theta is the angle of the force below the horizontal. Substituting the values into the equation, we get:
F_h = F_n * cos(35 degrees) = F_n * 0.819
Since the net force in the horizontal direction is zero, the horizontal component of the force must be equal to the force of friction. Therefore, we can write the equation:
F_h = F_f = 65 N
Substituting the value of the force of friction and the expression for the horizontal component of the force, we can solve for the magnitude of the force the nurse must exert:
F_n * 0.819 = 65 N
F_n = 65 N / 0.819 = 79.3 N
The nurse must exert a force of approximately 79.3 Newtons to move the cart at a constant velocity.
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An elevator is accelerating upward at a rate of 3.6 m/s2. a block of mass 24 kg hangs by a low-mass rope from the ceiling, and another block of mass 90 kg hangs by a low-mass rope from the upper block. (a) what are the tensions in the upper and lower ropes?
What eyeglasses would you prescribe for persons with the following conditions:
A near point of 56.6cm?
A far point of 56.6cm?
Final answer:
A person with a near point of 56.6cm is farsighted and needs convex lenses to correct their vision.
A person with a far point of 56.6cm is nearsighted and requires diverging lenses.
Explanation:
To correct for vision where a person has a near point of 56.6cm, they are likely farsighted. Farsightedness is corrected using convex lenses that converge the light before it hits the eye. This increases the eye's focusing power to bring nearby objects into clear vision.
For a person with a far point of 56.6cm, they are suffering from myopia, or nearsightedness. They require diverging lenses, typically negative diopter lenses, to spread out the light before it reaches the eye. This extends the far point of their vision, allowing them to see distant objects more clearly.
If the eyeglasses are held 1.50 cm from the eyes, we would need to consider the vertex distance in the lens formula to calculate the required lens power. The formula to determine the lens power needed, in diopters (D), is 1/f, where f is the focal length in meters. Since the eyeglasses are not in direct contact with the eye, the formula to find the effective focal length f' when considering vertex distance d is f' = f - d.
Imagine riding On a merry-go-round at the center. As you walk to the outer edge, the merry-go-round slows in order to conserve angular momentum. True or false?
A cart for hauling ore out of a gold mine has a mass of 435 kg, including its load. the cart runs along a straight stretch of track that is sloped 4.77° from the horizontal. a donkey, trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 438-n force for a distance of 193 m by means of a rope that is parallel to the ground and makes an angle of 13.3° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0167. use g = 9.81 m/s2. find the work that the donkey performs on the cart during this process.
A roller coaster car rapidly picks up speed as it Rolls down the slope. As it starts down the slope, its speed is 4 m/s. but three seconds later, at the bottom of the slope, its speed is 22 m/s what is the average acceleration?
A compact car has a maximum acceleration of 4.0 m/s2 when it carries only the driver and has a total mass of 1200 kg . you may want to review ( pages 108 - 109) . part a what is its maximum acceleration after picking up four passengers and their luggage, adding an additional 400 kg of mass?
Answer : The maximum acceleration will be 3.0 [tex] m/sec^{2} [/tex].
Explanation : We can calculate the force using the formula given below;
F = m . a ;
(F - Force; m - mass and a - acceleration);
given in the problem; mass (1) - 1200 kg; acceleration (1) - 4 [tex] m/sec^{2} [/tex]
So, on substituting we get,
F = [tex] m_{1} a_{1} [/tex] = 1200 X 4 = 4800 N; we get F = 4800 N
Now, when mass 400 kg is added then mass (2) will be 1600 kg and acceleration has to be found;
Here, the force remains the same; so it can be equated;
[tex] m_{1} a_{1} [/tex] = [tex] m_{2} a_{2} [/tex];
So, 1200 X 4 = (1200 + 400) x [tex] a_{2} [/tex]
Therefore, the answer will be [tex] a_{2} [/tex] = 3.0 [tex] m/sec^{2} [/tex]
How do single-celled decomposers get energy? A. They make their own food. B. They use a process called budding, C. They feed on living organisms. D. They break down dead organisms.
NEED ANSWER NOW GIVING 54 POINTS
Answer:
D
Explanation:
help me please in this
Elements with atomic numbers above 95 are called _____ elements
How long does it take a person at rest to breathe one mole of air if the person breathes 77.0 ml/s of air that is measured at 25 ∘c and 755 mmhg? express your answer numerically in seconds?
To determine how long it takes a person to breathe one mole of air at a rate of 77.0 ml/s, the ideal gas law and conversion to appropriate units would be necessary. However, the actual calculation was not provided, and thus the exact time in seconds cannot be given.
Explanation:The question asks about the time required for a person at rest to breathe one mole of air given a breathing rate of 77.0 ml/s at conditions of 25 ℃ and 755 mmHg. To solve this, we need to employ the ideal gas law. The ideal gas law relates the volume (V), pressure (P), temperature (T), and the number of moles (n) of an ideal gas to the universal gas constant (R).
We use the conditions given by the student to calculate the volume of one mole of air under the specified conditions. The ideal gas law is PV = nRT where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the universal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin.
First, convert the given pressure from mmHg to atm and the temperature from Celsius to Kelvin. Then solve for V when n is one mole.
However, without completing the calculation, we cannot provide the exact time needed as it depends on the precise volume one mole of air occupies at the given conditions. Once known, the time can be found by dividing this volume by the breathing rate (77.0 ml/s) and converting the result into seconds, which gives us the amount of time needed to breathe one mole of air.
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It would take approximately 318.2 seconds for a person at rest to breathe one mole of air, given the conditions of 25°C and 755 mmHg pressure. This calculation assumes ideal conditions and a constant breathing rate.
To determine how long it takes for a person at rest to breathe one mole of air, we need to calculate the volume of air that corresponds to one mole under the given conditions and then divide this by the breathing rate.
1. Calculate the volume of one mole of gas at STP (Standard Temperature and Pressure):
- Standard temperature (T): 0°C = 273.15 K
- Standard pressure (P): 1 atm = 760 mmHg = 101.325 kPa
Using the ideal gas law [tex]\( PV = nRT \)[/tex]:
\[ V_{\text{mole}} = \frac{nRT}{P} = \frac{(1 \text{ mol})(0.0821 \text{ L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1})(298.15 \text{ K})}{760 \text{ mmHg}} \approx 24.5 \text{ L} \][tex]\[ V_{\text{mole}} = \frac{nRT}{P} = \frac{(1 \text{ mol})(0.0821 \text{ L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1})(298.15 \text{ K})}{760 \text{ mmHg}} \approx 24.5 \text{ L} \][/tex]
2. Convert volume to milliliters (since the breathing rate is given in [tex]mL/s)[/tex]:
[tex]\[ V_{\text{mole}} = 24.5 \text{ L} \times 1000 \text{ mL/L} = 24500 \text{ mL} \][/tex]
3. Calculate the time to breathe one mole of air:
[tex]\[ \text{Time} = \frac{V_{\text{mole}}}{\text{Breathing rate}} = \frac{24500 \text{ mL}}{77.0 \text{ mL/s}} \approx 318.2 \text{ seconds} \][/tex]
Therefore, it would take approximately 318.2 seconds for a person at rest to breathe one mole of air, given the conditions of 25°C and 755 mmHg pressure. This calculation assumes ideal conditions and a constant breathing rate.
Using the mnemonic roy g. biv to remember the colors of the rainbow in the order of wavelength illustrates the use of
Final answer:
The use of the mnemonic ROY G BIV to remember the colors of the rainbow in the order of wavelength illustrates the concept of color vision in physics.
Explanation:
The use of the mnemonic ROY G BIV to remember the colors of the rainbow in the order of wavelength illustrates the concept of color vision in physics. Visible light consists of different wavelengths, and our eyes perceive these wavelengths as different colors. The mnemonic helps us remember the order of colors in the rainbow from longest to shortest wavelength: red, orange, yellow, green, blue, indigo, and violet.
For example, red light has a longer wavelength than orange light, and violet light has a shorter wavelength than green light. The mnemonic ROY G BIV is a helpful tool for remembering this sequence of colors and their corresponding wavelengths.
How many total atoms are in 0.250 g of p2o5?
Answer:
3.71 × 10²¹ atoms
Explanation:
The molar mass of P₂O₅ is 283.89 g/mol. The moles of P₂O₅ corresponding to 0.250 g of P₂O₅ are:
0.250 g × (1 mol/283.89 g) = 8.81 × 10⁻⁴ mol
In 1 mole of P₂O₅ there are 6.02 × 10²³ molecules of P₂O₅ (Avogadro's number). In 8.81 × 10⁻⁴ moles of P₂O₅, the molecules of P₂O₅ are:
8.81 × 10⁻⁴ mol × (6.02 × 10²³ molecule/ 1 mol) = 5.30 × 10²⁰ molecule
In 1 molecule of P₂O₅, there are 7 atoms (2 atoms of P and 5 atoms of O). In 5.30 × 10²⁰ molecules of P₂O₅, the number of atoms is:
5.30 × 10²⁰ molecule × (7 atom/1 molecule) = 3.71 × 10²¹ atom
You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. both strike the wall at speed v, but while the ball bounces off with no loss of speed, the clay sticks. what is the change in momentum of the clay and ball, respectively
#86 can an object be moving when its acceleration is zero? can an object be accelerating when its speed is zero?
You are interested in space travel and want to learn more facts about space travel. Which would be the best source of information
A similarity between tariffs and sanctions is that both are sometimes used to?
Answer:
“Punish other nations”, is the right answer.
Explanation:
Tariffs and sanctions are the two different ways to do the same job. Tariffs are taxes on imports, for instance, tariffs that China placed on U.S. corn, it implies that anyone who desires to buy the American corn has to pay more charges to get it. At the same time, "Sanctions" are means to punish a country by preventing the sale or the purchase of something of that country, for instance, U.S. sanctions on the oil from Iraq. Therefore, both the methods are employed to punish a nation and make it more difficult to make trading deals with that nation.
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body diagram shows the forces acting on the cart. The mass of the cart, to the nearest whole number, is _______ kg
Answer:
Mass of the cart = 146 kg
Explanation:
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.
The cart accelerates at 1.4 m/s² horizontally.
Horizontal force = Fcosθ = 250 cos35° = 204.79N
We have F = ma
Substituting
204.79 = m x 1.4
m = 146.28 kg = 146 kg
Mass of the cart = 146 kg
Answer:
146 kg
Explanation:
Correct on Edge 2021
What is the correct reference frame for a given situation?
Each degree in the Kelvin scale equals how many degrees on the Celsius scale?
A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation relates u and wint
ΔU = -Wint
Consdier the work of of interaction is W =m*g*h - equation -1
and the Potential energy U.
Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h
Now we will write the equation for a Change in Potential energy ΔU,
ΔU = Uf - Ui
= 0-m*g*h
ΔU = -m*g*h --Equation 2
Now compare the both equation
Wint = -ΔU
we can rewrite the above equation
ΔU = -W.
So our Answer is ΔU = -W. .
Why is a standard system of measurement important?
A. Scientists want to share measurements data that they can understand
B. Each country wants its own set of measurement standards.
C. Measurements are seldom shared with scientists in different countries
D. Scientists speak many different languages.
The correct answer is A. Scientists want to share measurements data that they can understand
Explanation:
In science, a standard system of measurement refers to a unified set or system of units that can be used internationally to measure data related to distance, temperature, energy, mass, etc. For example, the International System of Units establishes distance should be measured in meters rather than in inches, feet, etc. This guarantees all around the word scientist use the same units and can later share measurements despite language, culture or country because by using the same units scientists can understand data. Thus, a standard system of measurement is important because "scientists want to share measurements data that they can understand ".
A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that line makes a 20 degree angle relative to the horizontal. what is the tension in the line?
Answer : Tension in the line = 936.7 N
Explanation :
It is given that,
Mass of student, m = 65 kg
The angle between slackline and horizontal, [tex]\theta=20^0[/tex]
The two forces that acts are :
(i) Tension
(ii) Weight
So, from the figure it is clear that :
[tex]mg=2T\ sin\ \theta[/tex]
[tex]T=\dfrac{mg}{2\ sin\theta}[/tex]
[tex]T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}[/tex]
[tex]T=936.7\ N[/tex]
Hence, this is the required solution.
The tension in the line is [tex]\boxed{932.18{\text{ N}}}[/tex].
Further Explanation:
Free body diagram is the graphical representation or illustration of all the forces applied on the body with their directions. This helps us to visualize all the applied forces on the body with their directions and make us possible to solve complex problems based on kinematics.
Tension is the force exerted by the wire, cable or string like object when someone pulls it.
The force in the downward direction is due to gravity called as weight. Its magnitude depends upon the mass of the body and expressed as [tex]mg[/tex]. the term [tex]g[/tex] is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex].
Given:
The mass of the student is [tex]65\text{ kg}[/tex].
The sag angle with respect to the horizontal is [tex]20^\circ[/tex].
Concept:
The student walking in the slack line exerts tension on the slack line ties with two trees. There is tension on the both side of the slack line. The horizontal component of tension both side of the slack line will be canceling each other effects and the vertical component of tension both side of the slack line is balanced by the weight of the student.
It can be expressed as:
[tex]2T\sin\theta=mg[/tex]
Here, [tex]T[/tex] is the tension, [tex]\theta[/tex] is the sag angle, [tex]m[/tex] is the mass of object and [tex]g[/tex] is the acceleration due to gravity.
Substitute [tex]20^\circ[/tex] for [tex]T[/tex], [tex]65\text{ kg}[/tex] for [tex]m[/tex] and [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex] in the above equation.
[tex]\begin{aligned}2T\sin 20^\circ&=\left( {65{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right) \hfill\\T&=\frac{{\left({65{\text{ kg}}} \right)\left( {9.81{\text{ m/}}{{\text{s}}^2}}\right)}}{{2\sin 20^\circ}}\hfill\\&=932.18{\text{ N}}\hfill\\ \end{aligned}[/tex]
Therefore, the tension in the line is [tex]\boxed{932.18{\text{ N}}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
65 kg, student, walking, slackline, length, webbing, stretched, between, two, trees, sags, 20 deg, angle, relative, horizontal, tension, line, 932.18 N.
Which of the following most likely happens when the temperature of a chemical reaction is increased?
Molecules speed up?
Molecules stop reacting?
Reaction rate decreases?
Chemical reaction stops?
A molecules speed up because the temp is rising there for the molecules speed up
hope this helped =^-^=
(DUE IN 1 HOUR) A rocket of mass m is launched straight up with thrust F⃗ thrust.
a. Find an expression for the rocket's speed at height h if air resistance is neglected. Express your answer in terms of the variables Fthrust, m, h, and appropriate constants.
b. The motor of a 340 g model rocket generates 10 N thrust. If air resistance can be neglected, what will be the rocket's speed as it reaches a height of 86 m ?
From the statement, since rocket was launched so this means
that it start from zero, hence initial velocity is zero.
Since the rocket was launched vertically straight up, the
force acting on this motion is gravity.
A. The acceleration of the motion is then given
by:
a= F/m - g
Then we can use the general equation:
V^2 = Vo^2 + 2*a*h
where V is final velocity, Vo is initial velocity, a is acceleration,
h is height
Since we know that Vo = 0, so
V^2 = 2*a*h
V^2 = 2 (F/m – g) h
V = sqrt [2 h (F/m – g)]
B. Given that:
h = 86 m
F = 10 N
m = 340 g = 0.34 kg
Find for V:
V = sqrt [2 * 86 (10 / 0.34 – 9.8)]
V = 58.08 m/s
To find the rocket's speed at height h, use Newton's second law of motion and kinematic equations. The rocket's speed can be expressed as v = sqrt(2gh - 2Fthrusth/m). For the given values, the rocket's speed is 136.36 m/s.
Explanation:a. To find the rocket's speed at height h, we can apply the principles of Newton's second law of motion. Since air resistance is neglected, the only force acting on the rocket is the thrust force. The thrust force can be written as Fthrust = ma, where m is the mass of the rocket and a is the acceleration. The acceleration is given by a = g - g'' where g is the acceleration due to gravity and g'' is the acceleration due to the thrust force. Since g'' is directed opposite to the gravitational acceleration, we subtract it. On rearranging the equation, we get g'' = g -a. Substituting this into the equation Fthrust = ma, we obtain Fthrust = m(g - a). Solving for a, we get a = g - Fthrust/m. At height h, the rocket's speed can be found using the kinematic equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Since the initial velocity is 0, we have v² = 2as. Substituting the expression for acceleration, we get v² = 2gh - 2Fthrusth/m. Taking the square root of both sides, we have v = sqrt(2gh - 2Fthrusth/m).
b. Substituting the given values Fthrust = 10 N, m = 0.34 kg, h = 86 m, and g = 9.8 m/s² into the expression v = sqrt(2gh - 2Fthrusth/m), we can calculate the rocket's speed as it reaches a height of 86 m. Plugging in the values, we get v = sqrt(2 * 9.8 * 86 - 2 * 10 * 86 / 0.34) = 136.36 m/s.
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A small car with mass 0.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m. if the normal force exerted by the track on the car when it is at the top of the track (point
b.is 6.00 n, what is the normal force on the car when it is at the bottom of the track (point a)?
The normal force on a car traveling inside a vertical circular track differs at the top and bottom. At the top, the normal force is in the same direction as gravity, and at the bottom, it supports the car's weight against gravity plus provides the centripetal force. To find the normal force at the bottom, one must use the equations for circular motion taking into account the gravitational force.
Explanation:To determine the normal force on the car at the bottom of the track (point A), we must consider that the car is in circular motion, and at the top and bottom of the track, the forces acting on the car are different due to its position relative to the center of the circular path.
At the top of the track, the normal force and the weight of the car both act downwards. Since the car is in circular motion and is not accelerating vertically, the net force must be equal to the centripetal force required to keep the car moving in a circle. This can be represented by the equation N + mg = mv2/r at point B, where N is the normal force, m is the mass of the car, g is the acceleration due to gravity, v is the speed of the car and r is the radius of the circular path.
At the bottom of the track, the normal force acts upwards while the weight of the car acts downwards. The normal force at the bottom must be greater than at the top because it must support the car's weight in addition to providing the centripetal force. So, the equation at point A is N - mg = mv2/r.
From the information provided, we know that the normal force at the top is 6.00 N, the mass of the car is 0.800 kg, and the gravitational acceleration is 9.8 m/s2. By writing out both equations for the normal force at the top and bottom, and substituting the known values, we can solve for the unknown normal force at the bottom of the track.
Two children fight over a 200 g stuffed bear. the 25 kg boy pulls to the right with a 15 n force and the 20 kg girl pulls to the left with a 17 n force. ignore all other forces on the bear (such as its weight). at this instant, can you say what the acceleration of the bear is?
10 m/s² to the left.
Further explanationGiven:
The mass of a stuffed bear = 200 grams = 0.2 kg.The mass of a boy = 25 kg.The mass of a girl = 20 kg.A boy pulls to the right with a 15 N force.A girl pulls to the left with a 17 N force.Question:
What the acceleration of the bear is?
This is the case for implementing Newton's second law, i.e., 'the resultant force is proportional to the acceleration'. If we also use SI units then 'the resultant force is equal to the product of the mass and the acceleration'.
[tex]\boxed{\boxed{ \ F_{net} = m \cdot a \ }}[/tex]
The net force, i.e., [tex]F_{net}[/tex] is the resultant of all forces acting on an object, measured in newtons (kgm/s²).The mass measured is kilograms.The acceleration measured in m/s².Let's calculate the net force.
Referring to the horizontal axis marked a positive to the right, then
[tex]\boxed{ \ F_{net} = 15 - 17 = - 2 \ N \ }[/tex]
A negative sign indicates that the bear is moving to the left.
We continue to calculate the bear's acceleration. Remember, the mass that is applied is the mass of the bear that is accelerating.
[tex]\boxed{ \ F_{net} = m \cdot a \rightarrow a = \frac{F_{net}}{m_{bear}} \ }[/tex]
[tex]\boxed{ \ a = \frac{2 \ kg.ms^{-2}}{0.2 \ ms^{-2}} \ }[/tex]
Thus, the magnitude and direction of the acceleration is [tex]\boxed{\boxed{ \ a = 10 \ m/s^2 \ to \ the \ left \ }}[/tex]
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The acceleration of the bear is [tex]\boxed{10{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}[/tex] and its direction of motion is towards left side.
Further explanation:
The net force acting on the bear is equal to the difference between force applied by the girl on the bear and force applied by boy on the bear.
Given:
The mass of the bear is [tex]200{\text{ g}}[/tex].
The force applied by the boy on the bear, towards right side, is [tex]15{\text{ N}}[/tex].
The force applied by the girl on the bear, towards left side, is [tex]17{\text{ N}}[/tex].
The mass of the boy is [tex]25{\text{ kg}}[/tex].
The mass of the girl is [tex]20{\text{ kg}}[/tex].
Formula and concept used:
From the Newton’s law of motion,
[tex]\sum F=ma[/tex]
Here, [tex]\sum F[/tex] is the net force acting on the object, [tex]m[/tex] is the mass of object and [tex]a[/tex] is the acceleration.
The weight of the boy and the girl will not affect the net force applied by them on the bear.
The net force applied on the bear is equal to the difference between force applied by the girl and force applied by the boy because the forces applied by the boy and girl are parallel to each other but opposite in the direction.
Therefore, from the vector law, the resultant of the parallel vectors acting opposite in direction is equal to the difference between their magnitudes. The direction of the resultant can be specified as the direction of the vector on larger magnitude.
The expression for the net force on the bear is:
[tex]\boxed{ma={F_g} - {F_b}}[/tex] …… (1)
Here, [tex]{F_g}[/tex] is the force applied by the girl, [tex]{F_b}[/tex] is the force applied by the boy, [tex]m[/tex] is the mass of bear and [tex]a[/tex] is the acceleration of the bear.
Thus, the acceleration of the bear is [tex]\boxed{10{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}[/tex] and its direction of motion is towards left side.
Calculation:
Substitute the value of [tex]{F_g}[/tex], [tex]{F_b}[/tex] and [tex]m[/tex] in equation (1).
[tex]\begin{aligned}\frac{{200}}{{1000}}a&=17 - 15 \\0.2a&=2 \\a&=10{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}} \\ \end{aligned}[/tex]
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Answer detail:
Grade: College
Subject: Physics
Chapter: Dynamics of motion
Keywords:
Two children fighting, force on the bear, acceleration, newton's law, boy, girl, pulling bear towards left, 10m/s2, 10m/s^2, 10.0m/s2 , 10.0m/s^2, mass, net force, weight, magnitude and direction.
A particle travels 15 times around a 10-cm radius circle in 42 seconds. what is the average speed (in m/s) of the particle?
The average speed of the particle that travels 15 times around a 10 cm radius circle in 42 seconds is 0.224 m/s.
Explanation:To calculate the average speed of a particle, we use the formula: Speed = Distance ÷ Time. In this case, the distance travelled is the circumference of the circle (2πr) multiplied by the number of times the particle goes around the circle. Given that the particle travels 15 times around a 10 cm radius circle, the total distance is 2π*0.1m*15 = 9.42m. Since the time taken is 42 seconds, the average speed of the particle is then 9.42m ÷ 42s = 0.224 m/s.
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The particle's average speed is calculated by dividing the total distance travelled (15 times around the circle) by the total time taken (42 seconds). The speed is approximately 0.224 m/s.
Explanation:
The subject of your question is physics, specifically dealing with the concept of speed in circular motion. To solve this, we need to understand that the particle is traveling in a circular path with a known radius 15 times in a given time of 42 seconds. Therefore, the total distance travelled by the particle can be calculated using the formula for the circumference of a circle, which is 2πr (where r is the radius), multiplied by the number of circles made.
First, we need to convert the radius from cm to meters because the standard unit of measurement for speed is m/s. So, 10cm = 0.10m. Now, calculating the total distance travelled: 15 circles * 2 * π * 0.10m = 9.42m.
The speed of any object is defined as the ratio of the distance travelled to the time taken. So, the average speed of the particle will be calculated as (total distance / total time). Accordingly, the speed of particle would be 9.42m / 42s = 0.224m/s.
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what type of energy appears when a gymnast jumps on to a spring board? Question is on energy stores!Thanks
Answer:
gravity
Explanation: Because when you are jumping
Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at the point of ejection are 1 meter/second and 0.5 meters/second respectively, calculate the ratio of their horizontal distances of impact.
Final answer:
The ratio of the horizontal distances of impact for the two water jets is 2.
Explanation:
To calculate the ratio of the horizontal distances of impact, we need to first find the time of flight for each water jet. The time of flight can be determined using the formula:
time = distance / velocity
For the first water jet with a height of 50 centimeters and a velocity of 1 meter/second, the time of flight is:
time 1 = 0.5 meters / 1 meter/second
= 0.5 seconds
For the second water jet with a height of 100 centimeters and a velocity of 0.5 meter/second, the time of flight is:
time 2 = 1 meter / 0.5 meter/second
= 2 seconds
Now, we can calculate the horizontal distances of impact using the formula:
distance = velocity * time
For the first water jet, the horizontal distance of impact is:
distance 1 = 1 meter/second * 0.5 seconds
= 0.5 meters
For the second water jet, the horizontal distance of impact is:
distance2 = 0.5 meter/second * 2 seconds = 1 meter
Therefore, the ratio of their horizontal distances of impact is:
ratio = distance 2 / distance 1
= 1 meter / 0.5 meters
= 2
To practice problem-solving strategy 26.1 resistors in series and parallel. two bulbs are connected in parallel across a source of emf e = 11.0 v with a negligible internal resistance. one bulb has a resistance of 3.0 Ï , and the other is 3.0 Ï . a resistor r is connected in the circuit in series with the two bulbs. what value of r should be chosen in order to supply each bulb with a voltage of 2.4 v ?
To supply each bulb with a voltage of 2.4 V, the value of resistor r should be 0.6 ohms.
Explanation:To find the value of resistor r in order to supply each bulb with a voltage of 2.4 V, we can use the formula for voltage in a series circuit:
V = V1 + V2 + Vr
Since the voltage across each bulb should be 2.4 V, we have:
2.4 V = 2.4 V + 2.4 V + Vr
Simplifying the equation, we find that the value of resistor r should be 0.6 ohms.