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3 years ago

A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?

Physics

1 answer:

MakcuM [25]3 years ago

8 0

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2. This means that you have to multiply 5 by 5. 5 x 5= 25.

Edit: Also, because the surface is frictionless, it will make the object go faster too. Nothing can really slow it down unless something blocks it.

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An airplane flies with a velocity of 55.0 m/s [ 35° N of W] with respect to the air (this is

rodikova [14]

Answer:

21 m/s.

Explanation:

The computation of the wind velocity is shown below:

But before that, we need to find out the angles between the vectors

53° - 35° = 18°

Now we have to sqaure it i.e given below

v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°

v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951

v^2 = 440.6

v = √440.6

v = 20.99

≈ 21 m/s

Hence, The wind velocity is 21 m/s.

6 0

3 years ago

Question 8

Viefleur [7K]

Answer:

96 Joules

Explanation:

The formula for work is Fnet times displacement (F x d = w) which, in this case, 48N is the Fnet and 2m as the displacement. Then all we need to do is multiply these two and we get 96 Joules.

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3 years ago

A simple pendulum consists of a mass M attached to a string oflength L andnegligible mass. For this system, when undergoing smal

PilotLPTM [1.2K]

The frequency of the pendulum is independent of the mass on the end. (c)

This means that it doesn't matter if you hang a piece of spaghetti or a school bus from the bottom end. If there is no air resistance, and no friction at the top end, and the string has no mass, then the time it takes the pendulum to swing from one side to the other only depends on the length of the string.

8 0

3 years ago

10.A car is travelling at a constant speed of 27m/s. The driver looks away from the road for a 2.0s to tune in a station on the

Korvikt [17]

Explanation:

Distance = speed × time

d = (27 m/s) (2.0 s)

d = 54 m

5 0

3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

Explanation:

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

6 0

3 years ago