3 years ago

Which statements about acceleration are true?

Physics

1 answer:

bonufazy [111]3 years ago

4 0

It's definitely not B or C. There are things missing from A and D so we can't narrow it down any farther.

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An object moving at 30 m/s takes 5 sec. to come to a stop. What is the object’s acceleration

Klio2033 [76]

Answer:

6 m/s²

Explanation:

a=v/t

a= 30 m/s÷ 5 sec= 6 m/s²

4 0

3 years ago

A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr

MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

7 0

3 years ago

Read 2 more answers

The Answer to the question is plant cell. I got this wrong...Thanks Alot

Scilla [17]

This is a statement not a question .

7 0

3 years ago

A body of mass 500kg moving at a speed of 10m/s reaches the speed of 50m/s in 20s.The force exerted is

Leona [35]

Answer:

answer is 1000 N

formula used-

F= m x (v-u/t)

mark this as brainliest!

6 0

3 years ago

A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

3 years ago