You are facing North. Turn 90 degrees left. W.
Turn 180 degrees right. E.
Reverse direction. W.
Turn 45 degrees left. S.W.
Reverse direction. N.E.
In which direction are you now facing? North East
A 1040 kg car and 3360 kg truck undergo a perfectly inelastic collision. before the collision, the car was traveling southward at 1.80 m/s and the truck westward at 8.25 m/s. m/s. find the velocity (speed and direction) of the wreckage immediately after the collision.
Final answer:
To determine the wreckage's velocity after a perfectly inelastic collision, calculate the vector sum of the car's and truck's momenta, then divide by the total mass. Use Pythagorean theorem for the magnitude and arctan for the direction.
Explanation:
To find the velocity of the wreckage immediately after a perfectly inelastic collision, we apply the principle of conservation of momentum. Since the collision is inelastic, both vehicles stick together and move with a common velocity after the impact.
To calculate this, we will:
Determine the momentum of each vehicle before the collision.
Sum the momenta vectorially.
Divide the resultant momentum by the total mass of the system to find the velocity of the wreckage.
For the 1040 kg car and 3360 kg truck:
The momentum of the car is given by its mass times its velocity (1040 kg * 1.80 m/s south).
The momentum of the truck is given by its mass times its velocity (3360 kg * 8.25 m/s west).
Now we add these momenta vectorially, using the components in each direction:
Southward momentum (car's): 1040 kg * 1.80 m/s
Westward momentum (truck's): 3360 kg * 8.25 m/s
With the sum of momenta:
Total southward momentum = 1040 kg * 1.80 m/s = 1872 kg·m/s
Total westward momentum = 3360 kg * 8.25 m/s = 27720 kg·m/s
The magnitude of the wreckage's velocity can be found using Pythagoras' theorem:
√(1872^2 + 27720^2)
Finally, the direction of the velocity is given by the arctangent of the ratio between the southward and westward momentum components.
After calculating the values:
Magnitude of velocity: √(1872^2 + 27720^2) kg·m/s {÷} (Total mass 1040 kg + 3360 kg)
Direction of velocity: arctan(1872 kg·m/s {÷} 27720 kg·m/s)
This resultant velocity is the combined speed and direction of the wreckage after the collision.
The International Space Station is in orbit around the Earth 380 km above the surface. Which statement accurately describes its motion?
What property do the following elements have in common? sulfur, iodine, and magnesium A) Same phase at room temperature. B) Good conductors of electricity. C) Same number of valence electrons. Eliminate D) They form cations (positive ions).
i just did it and the answer us A
Primary action of the deltoid- 61)
Primary action of the adductor muscles 62)
Primary action of the erector spinae 63)
Primary action of the rectus abdomini
OPTIONs
A) pronation
B) rotation
C) dorsiflexion
D) flexion
E) adduction
F) circumduction
G) abduction
H) supination
I) extension
If the spheres are 19.6 meters above the ground, the time required for the aluminum sphere to reach the ground is
(1) 1s
(2) 2s
(3) 8s
(4) 4s
Answer:
(2) 2s
Explanation:
Remember that the time that it takes an object to fall from a certain distance is only determined by the force with which the object is pulled towards the center of the earth which is gravity, so any object with 0 velocity will drop at the same rate to the ground when dismissing resistance from the air, in to calculate this you just have to use the next formula:
[tex]H=\frac{g*t^2}{2}\\ t=\sqrt{\frac{2H}{g} }[/tex]
So we just insert the data that we have into the formula:
[tex]t=\sqrt{\frac{2H}{g} }\\t=\sqrt{\frac{2*19,6}{9,81} }\\t=\sqrt{4}\\ t=2 seconds[/tex]
In an elastic collision, an object with momentum 12 kg·m/s collides with another. the final momenta of each are 14 kg∙m/s and 16 kg∙m/s respectively. what was the initial momentum of the second object
We know that in an elastic collision the moments of the both the objects is conserved i.e
Intial momentum of object1 + Intial momentum of object2 =Final momentum of object1 + Final momentum of object2
Intial momentum of object1 =12 kg.m/s
Intial momentum of object2 = P (assume that P is the momentum)
Final momentum of object1= 14 kg.m/s
Final momentum of object2 =16 kg.m/s
On substituting all values we get
12 + P =14 kg.m/s + 16 kg.m/s
P = -18 kg.m/s
(-ve sign indicates the second object was moving in the opposite direction to the object1 before collision )
Therefore the initial momentum of the second object was 18 kg.m/s.
The greatest biodiversity on earth is found in the __________ biome.
A) taiga
B) grasslands
C) deciduous forest
D) tropical rainforest
The greatest biodiversity on earth is found in the tropical rainforest biome.
Explanation:Biome is actually another name for ecosystem. Rainforests are basically the wettest ecosystems and very diverse due to some reasons such as very high annual rainfall, high average temperatures, nutrient-poor soil, and high levels of biodiversity. Biomes or ecosystems are characterized by their climate and on that basis we can find which type of animals and plants can be found there. The greatest biodiversity on earth is found in the tropical rainforest biome.
The greatest biodiversity on earth is found in the tropical rainforest biome. Hence, option D is correct.
Biodiversity or biological diversity is the measure of variation at the species, genetic, and ecosystem levels. It comprises all the different kinds of life and supports life on Earth.
Biodiversity is important because it supports the entire life on the Earth including plants, animals, microorganisms, etc. Without biodiversity, a healthy ecosystem is not possible. Biome is called an ecosystem. Rainforest has the wettest ecosystems and has a higher annual rainfall and nutrient-rich soil.
The greatest biodiversity on earth is found in the tropical rainforests biome and hence, the ideal solution is option D.
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Which of these atoms is most likely to share electrons with other atoms?
It is the second one or the one with the letter F
If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s
2. Why was it important to examine both the color and the streak of your minerals? Think about streak and explain why it’s called a mineral’s “true color”. Answer in at least 2 sentences.
The examination of color and streak are the physical tests that are performed on the minerals to know their exact origin.
Two or more rocks or minerals can have similar color but they are unrelated in their origin and chemical composition. A color test can give false indication that two minerals or two rocks belong to the same type. A streak test can give direct indication of same or different origin as the mineral or rock is pressed against a hard substratum it liberates a powdery color residue which can be used to examine the exact color of the rock or mineral.Hence, this can be stated that a streak is useful to give true color of the mineral.
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What is the analogy heart:_____:stomach:digestive?
What are the four conditions needed to see an object?
Answer:
Four conditions need to be met for an object to be seen-an object, an eye, a source of light, and a direct, unblocked path between the object and the eye
hope this helps
Is the distance on a round-trip positive, negative, or zero?
The distance of every object in a round trip is always positive
Distance is a scalar quantity, the total distance of an object in any trip is the total path covered by the object from the starting point to the finish point.
In a round trip, the object start's from one point and makes a circular movement, then returns to the same starting point. The total distance of the object is the equivalent to the circumference of the circle. This measurement will be a positive value.On the other hand, the displacement of the object will be zero. Displacement is the change in the position of an object.
Thus, we can conclude that the distance of every object in a round trip is always positive while the displacement is zero.
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An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. Determine how long it takes to get to the maximum height of 24.0 m.
Answer:
It takes 2.04s to get to the maximum height
Explanation:
This is a vertical throw problem so it can be treated as a uniform accelerated rectilinear motion. For computing time we are going to use the formula:
[tex]v_{f}=v_{o}+g*t[/tex]
where[tex]v_{f}[/tex] is the final velocity, [tex]v_{o}[/tex] is the initial velocity, [tex]t[/tex] is the time and, [tex]g[/tex] is the gravity.
For solving this kind of problems we need at least three values. The values we have are:
[tex]v_{o} = 20\dfrac{m}{s}[/tex][tex]g = -9.8\dfrac{m}{s^{2}} [/tex] (negative because gravity's direction is oposite from the object's moving direction)[tex]v_{f}=0[/tex] (final velocity equals zero because at maximun height the object stops moving)Now:
[tex]v_{f}=v_{o}+g*t[/tex]
[tex]v_{f}-v_{o}=g*t[/tex]
[tex]\dfrac{v_{f}-v_{o}}{g}=t[/tex]
[tex]\dfrac{0-20}{-9.8}=t[/tex]
[tex]t=2.04s[/tex]
When monochromatic light is incident for a short period of time on a thin barrier having two slits, how will a screen placed in front of the slits appear?
Answer:
Interference pattern- alternate bright and dark fringes
Explanation:
When a monochromatic light is incident on a thin barrier having two slits, an interference pattern can be observed.
Dark and bright fringes would appear on the screen placed in front of it. It would not be random pattern. It would be a smooth pattern with distinctive dark fringe and bright fringe alternatively.
When the light from two source would superimpose, a constructive interference would lead to bright fringe and destructive interference would lead to dark fringe.
The Federal Communications Commission (FCC) has considered lifting the ban on in-flight cell phone use. This could allow people to have conversations on their cell phones during plane flights. Give your opinion. Should the FCC allow in-flight calls? Why or why not?
Answer:
I do not think the FCC should allow in-flight calls, because they would make the flight noisy and could make it difficult for passengers to hear the pilot or flight attendants.
Explanation:
The book slides from you to beth and then from beth to carlos, along the lines connecting these people. what is the work done by friction during this displacement?
The work done due to friction is [tex]\boxed{59.58{\text{ N}}}[/tex].
Further explanation:
Here, we have to calculate the total work done due to frictional force.
Frictional force is a non-conservative force and the work done by a non-conservative force is path dependent.
Given:
The mass of the physics book is [tex]1.9{\text{ kg}}[/tex].
Coefficient of friction between the book and floor is [tex]0.2[/tex].
The distance between observer that is (you) and Beth is [tex]8{\text{ m}}[/tex]
The distance between Beth and Carlos is [tex]8{\text{ m}}[/tex].
Formula and concept used:
The work done by the conservative force is equal to the product of force and the displacement.
Now, we will know about displacement,
Displacement: The shortest distance between the initial and final position of the object is known as displacement.
But, in case of the non-conservative forces, the work done is equal to the product of the force and the total distance travel by the object.
So, friction force can be calculated as,
[tex]{F_r} = \mu mg[/tex]
Here, [tex]m[/tex] is the mass of the book, [tex]\mu[/tex] is the coefficient of friction.
So, work done due to friction when observer slides the book to the Beth and Beth slides the book to Carlos.
[tex]\boxed{W = {F_r} \cdot d}[/tex]
Here, [tex]{F_r}[/tex] is the friction force, [tex]d[/tex] is the total distance travel against friction force.
[tex]\boxed{\begin{aligned}d&=8+8\\&=16\text{ m}\end{aligned}}[/tex]
Substitute the value of [tex]{F_r}[/tex] in above equation.
[tex]W = \mu mgd[/tex] …… (1)
Calculation:
Substitute [tex]0.2[/tex] for [tex]\mu[/tex], [tex]1.9{\text{ kg}}[/tex] for [tex]m[/tex] and [tex]16{\text{ m}}[/tex] for [tex]d[/tex] in equation (1).
[tex]\begin{aligned}W&=\left( {0.2} \right)\left( {1.9} \right)\left( {9.8} \right)\left({16}\right)\\&=59.58{\text{ N}}\\\end{gathered}[/tex]
The work done due to friction is [tex]\boxed{59.58{\text{ N}}}[/tex].
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Answer detail:
Grade: Senior School
Subject: Physics
Chapter: Work and Energy
Keywords:
Book slide, Beth, Carlos, work done, friction, four corners, square, 8m length, 59.58 N, connecting these people, displacement.
An object is placed so that the image formed is a real image of the same size as the object. What is the position of the object?
Two balls undergo inelastic collision. The y-momentum after the collision is 98 kilogram meters/second, and the x-momentum after the collision is 100 kilogram meters/second. What is the magnitude of the resultant momentum after the collision?
Answer:
D. 1.4 × 10^2 kilogram meters/second
This is one of the three main types of rocks one in which fossils are frequently found
if a snail starts at a position of 67cm and moves to a final position of 104 cm what is the displacement
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker is slipping with a velocity of 8.6 m/s north and the halfback is sliding with a velocity of 7.4 m/s east. What is the magnitude of the velocity at which the two players move together immediately after the collision
The two football players move together at a velocity of 5.99 m/s immediately after the collision. This value was obtained by conserving the momentum of the system: calculating individual momenta before the collision, summing them to obtain total momentum, and then dividing this by the total mass of both players. This is a typical conservation of momentum problem in high school level physics.
Explanation:The situation you're describing is a perfect example of a conservation of momentum problem in physics. In this problem, the two football players can be seen as a system, and their individual momenta before the tackle add up to yield the total momentum of the system after the tackle, when they're moving together.
Momentum is calculated as mass times velocity, so we first calculate the momentum of each player before the collision. For the linebacker: 120kg× 8.6m/s = 1032 kg×m/s north. For the halfback: 75kg× 7.4m/s = 555 kg×m/s east.
These two momentum vectors form a right triangle, with the hypotenuse representing the result vector or total momentum of the system after the tackle. We can use the Pythagorean theorem to calculate the magnitude of this vector: sqrt((1032²)+(555²)) = 1169 kg*m/s.
Since the players are moving together after the collision, the mass we use to find the final velocity should be the total mass of both players: 120kg + 75kg = 195 kg. The magnitude of the velocity at which the two players move together immediately after the collision can then be obtained by dividing the total momentum by the total mass: 1169kg×m/s / 195kg = 5.99 m/s.
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How is 6.3 written in scientific notation? 6.3 mc022-1.jpg 100 63 mc022-2.jpg 10–1 6.3 mc022-3.jpg 101 63 mc022-4.jpg 100
Final answer:
The number 6.3 is expressed in scientific notation as 6.3×10^0.
Explanation:
To express the number 6.3 in scientific notation, we need to follow a standard format where the number is between 1 and 10 followed by an exponent of 10.
For 6.3, this is already the case, so it can be expressed as 6.3×10^0 because any number raised to the power of 0 is equal to 1, so this does not change the value of 6.3.
Therefore, when writing in scientific notation, we are just acknowledging that we could move the decimal zero places and maintain the same value.
What is the resistance of a nichrome wire at 0.0 ∘c if its resistance is 200.00 ω at 11.5 ∘c?
We can use the temperature coefficient of resistance to determine the resistance of the nichrome wire at 0.0 °C. The temperature coefficient of resistance (α) is the amount by which the resistance of a material changes per degree Celsius of temperature change.
Given information:
Resistance of the nichrome wire at 11.5°C = 200.00 Ω
Temperature at which resistance is to be found = 0.0°C
We can use the following formula to find the resistance of the nichrome wire at 0.0°C:
R₂ = R₁ [1 + α (T₂ - T₁)]
Where,
R₁ = Resistance of the wire at temperature T₁
R₂ = Resistance of the wire at temperature T₂
α = Temperature coefficient of resistance
T₁ = Temperature at which R₁ is given
T₂ = Temperature at which R₂ is to be found
Since we are given the resistance of the nichrome wire at 11.5°C, we can take this as R₁ and T₁ as 11.5°C. We also know that the temperature coefficient of resistance for nichrome wire is 0.0004 per °C.
Substituting the given values into the formula, we get:
R₂ = 200.00 Ω [1 + (0.0004/°C) (0.0°C - 11.5°C)]
R₂ = 200.00 Ω [1 - 0.0046]
R₂ = 200.00 Ω (0.9954)
R₂ = 199.08 Ω
Therefore, the resistance of the nichrome wire at 0.0°C is 199.08 Ω.
Nichrome wire resistance at 0.0°C is 199.08 Ω.
Calculate nichrome wire resistance at 0.0°C using its temperature coefficient and resistance at 11.5°C.
The resistance R of a conductor at a temperature T is given by the formula:
[tex]\[ R_T = R_0 \left(1 + \alpha \Delta T\right) \][/tex]
Rearrange the formula to find [tex]\( R_0 \)[/tex] (resistance at 0.0°C):
[tex]\[ R_0 = \frac{R_{11.5}}{1 + \alpha \Delta T} \][/tex]
[tex]\( \Delta T \)[/tex]= 11.5 °C - 0.0°C = 11.5°C
Substitute the values:
[tex]\[ R_0[/tex]= 200.00Ω/(1 + (0.0004°C* 11.5°C))
[tex]\[ R_0[/tex] = 200.00Ω/(1 + 0.0046)
[tex]\[ R_0[/tex] = 200.00Ω/(1.0046)
[tex]\[ R_0[/tex] = 199.08 Ω
A moving 4.30 kg block collides with a horizontal spring whose spring constant is 223 n/m. the block compresses the spring a maximum distance of 5.00 cm from its rest position. the coefficient of kinetic friction between the block and the horizontal surface is 0.340. what is the work done by the spring in bringing the block to rest? submit answer tries 0/12 how much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? submit answer tries 0/12 what is the speed of the block when it hits the spring?
what is an exception to the rule that liquids are less dense than solids.
A 5-cm-external-diameter, 10-m-long hot-water pipe at 80°c loses heat to the surrounding air at 16°c by natural convection with a heat transfer coefficient of 25 w/m2·k. determine the rate of heat loss from the pipe by natural convection.
The rate of heat loss from a hot-water pipe by natural convection is calculated using the formula Q = h * A * ΔT. After plugging in the given values and conducting the appropriate calculations, the rate of heat loss turns out to be 2512 W.
Explanation:Let's determine the rate of heat loss from a hot-water pipe by natural convection. The formula to calculate the rate of heat loss though natural convection is: Q = h * A * ΔT, where:
Q is the rate of heat transfer h is the heat transfer coefficient, which in this case is 25 w/m2·k A is the surface area of the pipe, which we can calculate using A = π * d * l, where d is the diameter and l is the length of the pipe ΔT is the difference between the temperatures of the pipe and the surrounding air, which in this case is 80°c - 16°c = 64°c
Let's plug the numbers in:
First calculate the surface area, A = π * 0.05 m * 10 m = 1.57 m2. Then, to find Q, we use the formula Q = 25 w/m2·k * 1.57 m2 * 64 K = 2512 W. Therefore, the rate of heat loss from the pipe by natural convection is 2512 W.
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The rate of heat loss from the pipe by natural convection is 2512 W.
To determine the rate of heat loss from a hot-water pipe, we can use the formula :
Q = h ×A×ΔT
Where:
Q is the rate of heat transfer (W)h is the convective heat transfer coefficient (W/m²·K)A is the surface area (m²)ΔT is the temperature difference between the pipe surface and the surrounding air (K or °C)First, we calculate the surface area of the pipe:
The external diameter of the pipe is given as 0.05 m, and the length is 10 m. The surface area of a pipe is :
A = π ×D×L
Substituting the values:
A = π × 0.05 m × 10 m = 1.57 m²
Next, we calculating the temperature difference:
ΔT = T(pipe) - T(air) = 80°C - 16°C = 64°C
Finally, using heat transfer formula:
Q = h×A×ΔT = 25 W/m²·K × 1.57 m² × 64 = 2512 W
The rate of heat loss from the pipe by natural convection is 2512 W.
what factors affect the amount of solar energy that reaches earth's surface
Example 2: a horizontal cylindrical drum is 2.00 m in diameter and 4.00 m in length. the drum is partially filled with benzene (density = 0.879 g/cm3). what is the mass (kg) of benzene when the liquid depth is 0.85 m?
The mass (kg) of benzene is about 4470 kg
Further explanationThis problem is about Density.
Density is the ratio of mass to the volume of the object.
[tex]\large {\boxed {\rho = \frac{ m }{ V } } }[/tex]
ρ = density of object ( kg / m³ )
m = mass of object ( kg )
V = volume of object ( m³ )
Given:
Diameter of Cylinder = d = 2.00 m
Radius of Cylinder = R = d/2 = 2.00/2 = 1.00 m
Length of Cylinder = L = 4.00 m
Liquid Depth = H = 0.85 m
Density of Benzene = ρ = 0.879 g/cm³ = 879 kg/m³
Unknown:
mass of benzene = m = ?
Solution:
This problem is about Liquid Volume in Partially Filled Horizontal Tanks
Firstly we will calculate the volume of Benzene by using following formula:
[tex]V = A \times L[/tex]
[tex]V = ( \texttt{Area of Sector - Area of Triangle} ) \times L[/tex]
[tex]V = [ R^2 \cos^{-1}(\frac{R - H}{R}) - (R - H)\sqrt {(2RH - H^2)} ] L[/tex]
[tex]V = [ 1^2 \cos^{-1}(\frac{1 - 0.85}{1}) - (1 - 0.85)\sqrt {(2(1)(0.85) - 0.85^2)} ] 4[/tex]
[tex]V = [ \cos^{-1} (0.15) - 0.15 \sqrt{ 0.9775} ] 4[/tex]
[tex]V \approx \boxed {5.0877 ~ m^3}[/tex]
[tex]m = \rho \times V[/tex]
[tex]m = 879 \times 5.0877[/tex]
[tex]m \approx \boxed {4470 ~ kg}[/tex]
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Subject: Mathematics
Chapter: Density
Keywords: Temperature , Density , Iron , Sphere , Volume , Mass
What electric field strength is needed between the electrodes to achieve this deflection?