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3 years ago

A process occurs in which a system's potential energy increases while the environment does work on the system.

Physics

1 answer:

iogann1982 [59]3 years ago

4 0

Ok, now what do you want to know about it?

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an object has a constant acceleration of 3.2 m/s^2.at a certain instant its velocity is 10 m/s . what is its velocity after 5 se

aleksandrvk [35]

So here, you're looking for distance. The formula is D=vt+1/2at^2.

Lets plug in the informations.

10 m/s is our v (initial velocity)

5 second is out t(time)

3.2 m/s is our a(acceleration)

10m/s(5)+1/2(3.2m/s+5^2)

50m/s+1/2(28.2)

50m/s+14.1

Answer =64.1 m

Glad to help you out buddy. Let me know if you need help.

5 0

3 years ago

Objects appear different in size and shape in a container of water due to

o-na [289]

Answer:

A. refraction of light waves

Explanation:

Refraction happens when light travels from one medium to another and changes speed and bends. This also causes objects to look different sizes and shapes when they are submerged in water.

5 0

3 years ago

Based on the your prior knowledge, how far up a hill will a Hot Wheels car (0.01 kg) go if it is traveling at 5.0 m/s?

Sloan [31]

The formula for speed id speed is equal to distance/time

so to 5.0/0.01=500m/s.

8 0

4 years ago

What is the work done by a force equivalent to the weight of 100kg that moves an object 200km in 7.0min?

umka2103 [35]

Answer:

WD= 200 Joules

Explanation:

wd= f x d

wd=1000N x 0.2 m

wd=200 joules

5 0

3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers