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2 years ago

25.5 m/s in 5.75 s. What is the acceleration?

1 answer:

7 0

Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)

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The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.

vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of $q$ . Assume that $V$ is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

$\text{Potential Energy} = q \cdot V$ .

Keep in mind that since both $q$ and $V$ might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

$q = \rm -8.0\; \mu C$ ; (that's -8.0 microjoules, which equals to $\rm -8.0\times 10^{-6}\; J$ )
$V = \rm 40\; V$ .

Hence its electrical potential energy:

$\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J$ .

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this $\rm -8.0\; \mu C$ negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0

3 years ago

What is peer review?...

Bond [772]

<span>computing or networking is a distributed application architecture that partitions tasks or work loads between peers. Peers are equally privileged, equipotent participants in the application. They are said to form a peer-to-peer network of nodes.</span>

6 0

2 years ago

Read 2 more answers

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the

Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

$x = v_0 t \frac{1}{2} at^2$

Where,

x= Displacement

$v_0$ = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

$L = \frac{1}{2} a t_1 ^2$

For the second cart

$2L \frac{1}{2} at_2^2$

When the tenth car is aligned the length will be 9 times the initial therefore:

$9L = \frac{1}{2} at_3^2$

When the tenth car has passed the length will be 10 times the initial therefore:

$10L = \frac{1}{2}at_4^2$

The difference in time taken from the second car to pass it is 5 seconds, therefore:

$t_2-t_1 = 5s$

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

$\frac{1}{2} = (\frac{t_1}{t_2})^2$

$t_1 = \frac{t_2}{\sqrt{2}}$

From the relationship when the car has passed and the time difference we will have to:

$(t_2-\frac{t_2}{\sqrt{2}}) = 5$

$t_2 (\sqrt{2}-1) = 3\sqrt{2}$

$t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Replacing the value found in the equation given for the second car equation we have to:

$\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Finally we will have the time when the cars are aligned is

$18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2$

$t_3 = 36.213s$

The time when you have passed it would be:

$20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2$

$t_4 = 38.172$

The difference between the two times would be:

$t_4-t_3 = 38.172-36.213 \approx 2s$

Therefore the correct answer is C.

4 0

2 years ago

How to find impulse from mass and velocity?

olga nikolaevna [1]

The momentum change =mass*velocity change. But sincevelocity change is not known another strategy must be used to find the momentum change. The strategy involves first finding the impulse (F*t = 1.0 N*s). Since impulse = momentum change, the answer is 1.0 N*s.

7 0

3 years ago

A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m

kenny6666 [7]

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, $N - mg = F_c$

$N-mg=\frac{mv^2}{r}$

Now substitute the given values, we get

$N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}$

$N=15.68+46.08$

N = 61.76 N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

5 0

2 years ago