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3 years ago

Which equation can be used to solve for accelerarion

Physics

2 answers:

guapka [62]3 years ago

8 0

Answer:

a = $\frac{v-u}{t}$

MaRussiya [10]3 years ago

6 0

Answer:

a = Δv/Δt

Explanation:

The little triangle stands for change and t=time and v=velocity

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If a 580 N net force acts on a 40 kg what will the acceleration of the car be

myrzilka [38]

Answer:

I think u have to do 580N times 40KG

7 0

3 years ago

Read 2 more answers

What happens if :<br> . The test charge is not tiny.

docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

3 0

2 years ago

Please help with 9 & 10

Tanzania [10]

Add all the sec. and all the meter's and then add the meter's and sec. together

4 0

3 years ago

Consider the graph below that represents the variation of the velocity with respect to time of an object moving along the x - ax

vazorg [7]

In a velocity-time graph, the area under the curve represents the distance.

The distance traveled from 10s to 18 s is

$\begin{gathered} d\text{ = }\frac{1}{2}\times10\times(18-10) \\ =\text{ 40 m} \end{gathered}$

Final Answer: The distance traveled is 40 m from time 10 s to 18 s.

3 0

1 year ago

Two objects exert a gravitational force on 8N on one another what would that force be if the mass of both objects were doubled?

jolli1 [7]

Gravitational force of attraction between two objects can be calculated from the following formula:
$F = \frac{m_1 \cdot m_2}{r^{2}}$
Where m with subscript stands for mass of the object and r is the distance between them.
When we double the mass of those two objects, distance between them stays the same, while in the numerator we have:
$2\cdot m_1 \cdot 2 \cdot m_2 = 4\cdot m_1 \cdot m_2$
When numerator in the second case 4 times greater than the 'original' numerator and denominator stays the same, force becomes 4 times greater.

3 0

3 years ago