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3 years ago

How do you find the change in velocity

Physics

1 answer:

arlik [135]3 years ago

6 0

Answer:

Multiply the acceleration by the time to obtain the velocity change.

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What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?

givi [52]

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

6 0

2 years ago

If a person pushes a box up a ramp, in which direction will the box exert a force on the person? up the ramp toward the person d

EleoNora [17]

Down the ramp toward the person

4 0

3 years ago

Read 2 more answers

A stone takes 5.4 seconds to fall from the top of a cliff. The cliff is

Allushta [10]

Answer:

143

Explanation:

Using one of the 3 fundamental equations in physics, y=vo*t+1/2gt^2, we can use this equation to find the total distance that was traveled.

Acceleration due to gravity is always 9.8m/s^2 and time is 5.4s, we also have no initial velocity.

Given this, we can plug in the known variables.

y=0t+1/2*9.8*5^2

simplify,

y=4.9*5.4^2

y=4.9*29.16

y=142.884m which we can round up to 143 meters

Final Answer: 143 meters

4 0

2 years ago

A car drives 40 km due east and then 50 km due west .what is the cars overall displacement?

Firdavs [7]

Answer:10 km westExplanation:he go 40 east then 50 west 50-40 is 10 so he displaces 10 km and as west is more than east in terms of km so we will say that it's 10 km west pls mark as brainliest thanks

7 0

3 years ago

Read 2 more answers

If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.

Schach [20]

Answer:

The magnitude of each charge is $2.82\times10^{-6}\ C$

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

$F=\dfrac{kq^2}{r^2}$

Where, q = charge

r = separation

Put the value into the formula

$20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}$

$q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}$

$q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}$

$q=2.82\times10^{-6}\ C$

Hence, The magnitude of each charge is $2.82\times10^{-6}\ C$

6 0

3 years ago