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2 years ago

A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the

Physics

1 answer:

Nitella [24]2 years ago

4 0

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

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A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$

Substitute the given values and solve the speed;

$v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s$

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

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7 0

2 years ago

If we cannot see the whole society what can we see

creativ13 [48]

I believe it's People interacting. I hope i've helped you! (:

8 0

2 years ago

The SAF operates the M113 Ultra APC.

HACTEHA [7]

The volume of the object must be no larger than $11.15 m^3$ .

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

$\rho = 1000 kg/m^3$

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

$\rho = \frac{m}{V}$

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

$m=1.115\cdot 10^4 kg$

Therefore, we can re-arrange the equation to find its volume:

$V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3$

So, the volume of the object must be no larger than $11.15 m^3$ .

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3 years ago

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where n=1,2,3 or ...n

L=Length of string

T=Tension

$\mu =$ Mass per unit length

When string is clamped at mid-point

Effecting length becomes $L'=0.5 L$

Thus new Frequency becomes

$f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}$

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz

4 0

2 years ago

ASAP WILL GIVE BRAINLIEST Take a piece of paper, some nails and a magnet bar, place a magnet at the centre of the paper, and spr

Zanzabum

Answer: find the answer in the explanation

Explanation:

When a magnet is placed at the centre of the paper, and the nails are sprinkled on the paper, what will happen to the nails is that, the nails will form a pattern on the paper according to the magnetic field of the bar magnetic pole.

Other phenomena you can observe are:

1.) The nails will align themselves and show some lines of forces which is equivalent to the magnetic field lines

2.) The direction of the line of forces

3.) The strength of the magnetic field pole.

7 0

2 years ago