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3 years ago

Explain the importance of wearing your seatbelt using Newton’s laws and motion

Physics

1 answer:

Elis [28]3 years ago

7 0

Explanation:

Newton's first law of motion states: "an object in motion stays in motion, and an object at rest stays at rest, until acted upon by an unbalanced force."

When you're riding in a vehicle, your body is moving forward. When the vehicle slows or turns, your body's inertia continues to move you forward. The seatbelt is the unbalanced force that pushes you back into your seat. Without it, you would slide off your seat.

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High altitude high velocity rivers of air are called

forsale [732]

The answer to the question would be jet streams.

3 0

3 years ago

Read 2 more answers

Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25

Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: $d_{1x} = 5.2 cos 25^{\circ}=4.7 mi$

North: $d_{1y} = 5.2 sin 25^{\circ}=2.2 mi$

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: $d_{2x}=0$

North: $d_{2y} = 5.0 mi$

So the components of the total displacement are

East: $d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi$

North: $d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi$

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

$d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi$

5 0

3 years ago

Write the reason behind the atmosphere does not escape away from the earth

postnew [5]

Answer:

i knew it but now i dont

Explanation:

4 0

3 years ago

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Speed of a car is 30 m/s. Find its speed in km/h.

katovenus [111]

Answer:

108km/h

Explanation:

1km=1000m

3600s=1 hr

M/s= 3600/1000km/hr= 3.6km/hr

30m/s= 30×3.6km/hr = 108km/hr

4 0

3 years ago

A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dista

Vitek1552 [10]

(a) $y(t)=250 - 4.9 t^2$

For an object in free-fall, the vertical position at time t is given by:

$y(t) = h + ut - \frac{1}{2}gt^2$

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

$y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2$

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

$y(t) = h - \frac{1}{2}gt^2=0$

So for the stone in the problem, we have

$250 - 4.9 t^2 = 0$

Solving for t, we find:

$t=\sqrt{\frac{250}{4.9}}=7.14 s$

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

$v(t) = u - gt$

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

$v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s$

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

$y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2$

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

6 0

2 years ago