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adell [148]
2 years ago
11

how long would it take for a resultant upward force of hundred Newton to increase the speed of 50 kg object from hundred metre p

er second to 150 metre per second?​
Physics
1 answer:
Hunter-Best [27]2 years ago
5 0

To answer this question, two equations are needed.

F = m * a (force = mass * accelaration)

Vf =Vi + (a * t) (final velocity = initial velocity + (accelaration * time)

Knowing that the force is 100 N and the mass is 50 kg we can use the first equation to solve for the acceleration which would be 2 m/s^2.

Plugging in all we know to the second equation we get:

150 m/s = 100 m/s + (2 m/s^2 * t)

This gives us t = 25 seconds

You might be interested in
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
-A 180 kg hippo is riding a bicycle at a speed of 6.0
Vlad1618 [11]

Answer:

0.0675 seconds

Explanation:

From the question,

We apply newton's second law of motion

F = m(v-u)/t.................... Equation 1

Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.

make t the subject of the equation

t = m(v-u)/F................... Equation 2

Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)

Substitute these value into equation 2

t = 180(0-6.0)/-1600

t = -1080/-1600

t = 0.0675 seconds.

8 0
2 years ago
An object can have forces acting upon it, but might not accelerate<br><br> True<br> or<br> False?
vivado [14]

Answer:

true

Explanation:

if you apply force to the top of a square it will not move

8 0
3 years ago
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow
allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

l = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

l = l1 + l2 + 2√l1l2cos(∅)

where

l1 = 2.25 mW

l2 = 2.025 mW

∅ = π

so we substitute

l = 2.25 + 2.025 - 2√(2.25 × 2.025)

l = 4.275 - 4.26907

l = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer  

5 0
2 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
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