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Vsevolod [243]
3 years ago
8

An object is moving along a straight line. The graph shows the velocity of the object over

Physics
2 answers:
iren [92.7K]3 years ago
8 0

Answer:

From 9 to 10 sec the velocity is decreasing at a constant rate.

Change in V = 20 m/s - 30 m/s = -10 m/s

Since acceleration is constant from point E to point F (straight line)

and t = 9.5 sec is on this line

a = (20 m/s - 30 m/s) / 1 sec = -10 m/s^2  

IRISSAK [1]3 years ago
4 0

Answer:

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Explanation:

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A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.
stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

v^{2} = v^{2}_{0}  + 2ax

You plug into the equation to get:

0 = 15^{2} + 2a(14)

solve for a to get

-8.04 m/s2

3 0
3 years ago
A 5.00 kg crate is on a 21.0° hill.
mojhsa [17]

Answer: 4575N

Explanation:

For y component, W = mgcosø

W = 500×9.8cos21

W = 4574.54N

Find the diagram in the attached file

8 0
4 years ago
A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
77julia77 [94]
C because when the part gets out of the probe it would no longer stay contacted
6 0
3 years ago
On Venus, the acceleration due to gravity is 8.87 m / s 2 . How far would a 17 g rock fall from rest in 6.5 s if the only force
algol [13]

Answer:

187.38 m

Explanation:

Using the equation of motion

s = ut + 1/2gt²...................... Equation 1

Where s = distance of fall, u = initial velocity of the rock, t = time taken for the rock to fall from rest, g = acceleration due to gravity of venus.

Given: u = 0 m/s ( from rest), t = 6.5 s, g = 8.87 m/s².

substituting into equation 1

s = 0(6.5) + 1/2(8.87)(6.5)²

s = 0 + 374.7575/2

s = 187.38 m.

Hence the rock will fall 187.38 m

7 0
4 years ago
Which of the following could be used to create an open circuit?
Arturiano [62]
A switch
What are the answers choices
4 0
3 years ago
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