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3 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]3 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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When a solid uniform sphere is spinning about an axis of rotation through its center, its rotational kinetic energy is K and mom

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What is the magnitude and direction (right or left) of the

Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

$Fy_{net}=F_{n}+F_{g}$ (1)

Where $F_{n}=12 N$ is the Normal force, directed upwards and $F_{g}=-12 N$ is the weight of the box (gravity force), directed downwards.

$Fy_{net}=12 N-12 N$ (2)

$Fy_{net}=0 N$ (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

$Fx_{net}=F_{left}+F_{right}$ (4)

Where $F_{left}=-3 N$ and $F_{right}= 15 N$

$Fx_{net}=-3 N + 15 N$ (5)

$Fx_{net}=12 N$ (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

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3 years ago

How is property belonging to individuals protected under the Fifth Amendment?

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By preventing the government from taking property without fair payment.

Explanation:

Answer 1

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Which statement is true about tomato juice? A) It is an acid. B) It has a pH of 7.0. Eliminate C) It is strongly basic in nature

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In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls

sweet-ann [11.9K]

Answer:

Acceleration= $0.5m/s^2$

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration= $a=\frac{F}{m}$

Using the formula

Acceleration of box= $\frac{20}{40}=0.5m/s^2$

The acceleration of the box= $0.5m/s^2$

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

$a=\frac{30}{40}=\frac{3}{4} ms^{-2}$

$v^2-u^2=2as$

Substitute the values

$v^2-0=2\times \frac{3}{4}\times 0.3=0.45$

$v^2=0.45$

$v=\sqrt{0.45}=0.67m/s$

Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s

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3 years ago