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slavikrds [6]
3 years ago
12

You travel down the highway at a steady rate of 75 mph (33.53 m/s)  for a total of 25 minutes, calculate how far you traveled i

n meters
Physics
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

50280 meters

Explanation:

33.52 meters/seconds is 2011.2 meters/minutes (multiply by 60)

2011.2 meters/m * 25 minutes = answer

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What is the displacement of the GBS cross-country team if they begin at the school, run 10 miles, and finish back at the school?
olga2289 [7]

Answer:

0

Explanation:

8 0
3 years ago
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g
REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

3 0
3 years ago
From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

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Which of the following is an example of kinetic energy being converted to potential energy?
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Of the following...?? Is there more to this question? :)
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What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
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Answer:

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Explanation:

7 0
3 years ago
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