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2 years ago

10

If a student is walking against the wind, would the student be traveling faster or slower than a person walking with the wind an

d why?

1 answer:

Ivahew [28]2 years ago

5 0

Answer:

Slower

Explanation:

Slower because the air would be pushing you back, and the other person would be walking faster because they are being pushed forward by the wind

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A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on M

Pie

Answer:

Gravitational field strength =weight/mass

Explanation:

14.8N/4.0kg

3.7N/kg

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2 years ago

A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

JulsSmile [24]

$m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}$

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2 years ago

A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

Olenka [21]

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

7 0

3 years ago

Read 2 more answers

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Ksenya-84 [330]

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3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

9 months ago