"on a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. when max pulls on the sled with 12 n of force, directed at an angle of 15° above the ground, how much work does max do on the sled as he pulls his sister 25 m in the snow?"

Final answer:

The airplane's acceleration that caused the passenger to feel 170 N heavier is 2.33 m/s² upwards.

Explanation:

When a person experiences an increase in weight due to airplane turbulence, it indicates that there's an additional force acting on them because of the airplane's acceleration. To find the magnitude and direction of the airplane's acceleration, we can use Newton's second law of motion (F=ma), where F is the force, m is the mass, and a is the acceleration.

The student feels 170 N heavier, implying that the upward acceleration of the airplane is causing this additional force. Since the student's mass is 73 kg, we can calculate the acceleration using the formula a = F/m, where F is the additional force (170 N) and m is the mass of the student (73 kg). Hence, a = 170 N / 73 kg = 2.33 m/s². This is the magnitude of the acceleration.

The direction of the acceleration is upwards, as it increases the normal force which is felt as an increase in apparent weight.

Answer: Option (d) is the correct answer.

Explanation:

Net force is defined as the sum of total number of forces acting on a substance.

For example, when you and your friend are pushing a door in order to slide it open then it means that the net force is sum of force applied by you and force applied by your friend.

Therefore, the force applied earlier was less but it increase when your friend also started applying force.

Thus, we can conclude that if your friend stands behind you and helps you push, then the net force applied will increase.

Final answer:

The largest torque that can be applied to the solid aluminum shaft is 1.5 MPa.

Explanation:

To determine the largest torque that can be applied to the solid aluminum shaft, we need to consider the allowable shear stress and the diameter of the shaft. The formula to calculate torque is t = rF sin 0, where t is the torque, r is the radius, F is the applied force, and 0 is the angle between the force and the radius. In this case, since the force is perpendicular to the radius, the angle is 90 degrees, so sin 0 = 1.

The diameter of the shaft is given as 50mm, which means the radius is 25mm. We need to convert the radius to meters, so the radius is 25/1000 = 0.025m. The allowable shear stress is given as 60MPa.

Using the formula for torque, t = rF = (0.025m)(60MPa) = 1.5MPa.

Therefore, the largest torque that can be applied to the shaft is 1.5MPa.

Answer:

Acceleration, [tex]a=-6.705\ m/s^2[/tex]

Explanation:

It is given that,

Velocity of the car, u = 60 mph = 26.82 m/s

Finally it comes to stop, v = 0

Time taken, t = 4 s

We need to find the acceleration of the car. The change in velocity divided by time is called the acceleration of the object. Mathematically, it is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{0-26.82}{4}[/tex]

[tex]a=-6.705\ m/s^2[/tex]

Negative sign shows the car is decelerating. So, the acceleration of the car is [tex]6.705\ m/s^2[/tex]. Hence, this is the required solution.

The car's original speed can be calculated by utilizing the kinematic equation which is a principle in physics. Given that the car came to a stop with constant deceleration in a certain distance, the estimated initial speed was approximately 26.9 m/s.

The subject of the question involves concepts in physics specifically related to the equations of motion. To solve this, we can utilize the kinematic equation which relates initial velocity, acceleration, and distance: v2 = u2 + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is acceleration and 's' is distance. Since the car comes to a stop, our final velocity v = 0 m/s. We know that the distance s = 30m and the acceleration is given as a = -3.50 m/s2 (negative because it's deceleration). From these, we can find the initial velocity 'u'. When we substitute these values into the equation, we can solve for 'u' as √(v2 - 2as), which equals approximately √(0 - 2*-3.50*30) ≈ 26.9 m/s. Therefore, the car's original speed was approximately 26.9 m/s.

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The correct answer would be letter d, 8.3.

Solution for the problem follows:

Given are:

B = 6i - 8j 

A is unknown; let A be = mi + nj 

A+B is along the x axis (therefore A+B = Ki + 0j, where K is unknown,

but then again the magnitude of A+B is the similar as the magnitude of A,

so mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 

A+B, from simple vector addition, will be now (m+6)i + (n-8)j.

Ever since we previously know A+B = Ki + 0j, we now know that: 

m+6 = K 

n-8 = 0, which implies n=8. 

Thus, K^2=m^2+n^2 ====> (m+6)^2 = m^2 +8^2 

= m^2 + 12m + 36 = m^2 + 64 

= 12m = 28 

= m = 2.33333... 

Therefore, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333

The magnitude of [tex]\vec A[/tex] is [tex]8.3[/tex].

Further explanation:

A vector is a quantity having magnitude and direction both. It is represented as the product of magnitude and direction vector.

Given:

The vector is given as [tex]\vec B = 6\hat i - 8\hat j[/tex].

The direction of resultant vector is in the X- direction.

Concept used:

Consider a vector [tex]\vec A = a\hat i + b\hat j[/tex] which is added to [tex]\vec B[/tex] to get a resultant vector [tex]\vec C[/tex]. The resultant vector is directed in the positive X-direction which means that the y- component of the vector is zero.

The expression for the resultant vector is given as.

[tex]\vec C = \vec A + \vec B[/tex]

Substitute [tex]6\hat i - 8\hat j[/tex] for [tex]\vec B[/tex] in the above expression.

[tex]\begin{gathered}\vec C = \left( {a\hat i + b\hat j} \right) + \left( {6\hat i - 8\hat j} \right) \\= \left( {a + 6} \right)\hat i + \left( {b - 8} \right)\hat j \\ \end{gathered}[/tex]

The resultant vector is represented as.

[tex]\vec C = x\hat i + 0\hat j[/tex]

Compare the above two expression of resultant vector.

[tex]x = a + 6[/tex]                                         …… (1)

[tex]\begin{aligned}0&=b-8\\b&=8\\\end{aligned}[/tex] 

The magnitude of resultant vector is equal to the magnitude of [tex]\vec A[/tex].

The expression for the magnitude of [tex]\vec A[/tex] is given as.

[tex]\left| {\vec A} \right| = \sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}[/tex]                           …… (2)

The expression for the magnitude of [tex]\vec C[/tex] is given as.

[tex]\left| {\vec C} \right| = \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]

Compare the above two expressions.

[tex]\sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}= \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]

Substitute 8 for [tex]b[/tex]  in above expression.

[tex]\begin{aligned}\sqrt {\left( {{a^2}} \right) + \left( {{8^2}} \right)}&=\sqrt {{{\left( {a + 6} \right)}^2}}\\{a^2} + 64&={a^2} + 36 + 12a \\12a&=28 \\a&=2.33 \\ \end{aligned}[/tex]

Substitute 2.33 for a and 8 for b in equation (2).

[tex]\begin{aligned}\left| {\vec A} \right|&=\sqrt {{{\left( {2.33} \right)}^2} + {{\left( 8 \right)}^2}}\\&=8.33 \\ \end{aligned}[/tex]

Thus, the magnitude of vector A is [tex]8.33[/tex].

Learn more:

1.  Motion under friction https://brainly.com/question/7031524.

2.  Conservation of momentum https://brainly.com/question/9484203.

3. Circular motion https://brainly.com/question/9575487.

Answer Details:

Grade: College

Subject: Physics

Chapter: Vectors

Keywords:

Vectors, product, magnitude, direction, resultant vector, adding vector, subtraction of vector, 2.33, 8, 8.33, 8.3, 8.33.

Final answer:

To find Neil's velocity relative to Barbara, their individual velocities are considered as vectors and subtracted. Neil's relative velocity is calculated to be approximately 6.74 m/s at 69° north of west.

Explanation:

To determine Neil's velocity relative to Barbara, we need to consider both Neil’s and Barbara's velocities as vectors and subtract the velocity of Barbara from Neil's velocity. Barbara is moving due south at 6.3 m/s, and Neil is moving due west at 2.4 m/s. By defining south as the negative y-direction and west as the negative x-direction, we can represent Barbara’s velocity vector as (0 m/s, -6.3 m/s) and Neil's as (-2.4 m/s, 0 m/s).

To find Neil’s velocity relative to Barbara, we calculate the vector difference: Neil’s velocity minus Barbara's velocity:

VNB = VN - VB = (-2.4 m/s, 0 m/s) - (0 m/s, -6.3 m/s) = (-2.4 m/s, 6.3 m/s)

To find the magnitude of Neil's relative velocity (VNB), we use the Pythagorean theorem: |VNB| = √((-2.4 [tex]m/s^2[/tex] + (6.3 [tex]m/s^2[/tex]), which gives:

|VNB| = √(5.76 + 39.69) = √(45.45) ≈ 6.74 m/s

The direction relative to due west can be found using the arctangent of the y-component over the x-component:

θ = arctan(6.3 / 2.4) ≈ arctan(2.625) ≈ 69°

Therefore, Neil's velocity relative to Barbara is approximately 6.74 m/s at an angle of 69° north of west.

The pressure at the stagnation point is 149.854 lb/ft².

According to Bernoulli's equation, the total pressure at a given point is the sum of the static pressure and the dynamic pressure. The dynamic pressure is given by 0.5 * ρ * v^2, where ρ is the density of the fluid and v is the velocity of the fluid. In this case, the static pressure is -2.0 psi (a vacuum) and the velocity is 150 ft/s. The density of air at sea level is approximately 1.14 kg/m³, which is equivalent to 0.075 lb/ft³. Converting the units, we have -2.0 psi = -18.896 lb/ft². Plugging in the values into Bernoulli's equation, we can calculate the dynamic pressure:

Dynamic pressure = 0.5 * (0.075 lb/ft³) * (150 ft/s)^2 = 168.75 lb/ft².

The total pressure at the stagnation point is equal to the sum of the static pressure and the dynamic pressure. Therefore, the pressure at the stagnation point is:

Total pressure = -18.896 lb/ft² + 168.75 lb/ft² = 149.854 lb/ft².

Answer:B,C, They cannot tell the difference between people and other objects. & They will wait for crossing pedestrians even if signaled to keep driving.

Explanation: i just did it on edg

Driverless cars can disrupt traffic due to slow speeds and over-caution, struggle to differentiate between people and objects, unexpectedly move to avoid hazards, and may not allow human control in emergency situations.

Driverless cars, or autonomous vehicles, have several disadvantages that might affect their performance and usability. Firstly, these vehicles often operate at safer, slower speeds than typical road traffic, which can disrupt normal traffic flow. Secondly, while they use advanced sensors to detect their surroundings, they might struggle to differentiate between people and other objects. This could potentially lead to harmful incidents. Additionally, they might be overly cautious and halt for crossing pedestrians even when it's unnecessary, causing further disruption.

Moreover, unpredictable actions like veering off the road to circumvent hazards could pose risks to passengers or other vehicles. Lastly, a key issue is if these vehicles don't permit the human driver to override control manually in crisis situations, which can decrease the human's ability to avert dangerous situations.

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Final answer:

The roller coaster's speed at the top of the loop is calculated using the centripetal acceleration formula, with the given downward acceleration of 1.50 g which equals 14.7 m/s² and the radius of curvature of 11.0 m to find the velocity.

Explanation:

The speed of the roller coaster at the top of the loop can be determined by utilizing the concept of centripetal acceleration. Centripetal acceleration is provided by the gravitational force when the roller coaster is at the top of the loop. Given that the downward acceleration is 1.50 g, and knowing that 1 g equals 9.8 m/s2, we can calculate the centripetal acceleration as 1.50 times 9.8 m/s2. The formula for centripetal acceleration (ac) is ac = v²/r, where v is the velocity and r is the radius of curvature. Rearranging the formula to solve for v gives us v = √(ac × r). Plugging in the values, we have v = √((1.50 × 9.8 m/s2) × 11.0 m), which yields the speed of the roller coaster at the top of the loop.

The direction of the centripetal force when applied to an object is perpendicular to the object's motion.

Any motion along a curved road is accelerated, necessitating the application of force to the path's center of curvature. This force is known as the centripetal force, which is a force that seeks its center.

The direction of centripetal force, or even net force, is always that of acceleration. As an illustration, consider how quickly the earth is moving toward the sun. We are not moving toward the sun since the acceleration vector for planet Earth is constantly pointing in that direction and the velocity is perpendicular to it.

When force applied to an item, the centripetal force's direction is perpendicular to the object's motion.

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The horizontal force would be necessary to displace its position 0.16 m to one side will be 10.7 N.

Horizontal force is defined as a force applied in a direction parallel to the horizon. A force exerted on a body has two components: a horizontal component and a vertical component.

Given that,

m = 30 kg,

W = mg = (30 kg)*(9.8 m/s²)

W = 294 N, the weight of the chandelier.

Let T = tension in the wire.

Let F = the horizontal force required to displace the chandelier by 0.16 m.

The angle that the wire makes with the vertical is given by:

sin(θ) = 0.16/4.4 = 0.0364

θ = sin⁻¹ 0.0364

θ = 2.084°

For equilibrium,

T cos(θ) = W

0.9993T = 294

T = 294.195 N

We know that,

F = Tsin(θ)

F = 294.195*0.0364

F = 10.709 N

Thus, The horizontal force would be necessary to displace its position 0.16 m to one side will be 10.7 N.

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