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podryga [215]
3 years ago
5

An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppos

e the mass of the object is doubled, and the magnitude of the force acting on it is quadrupled.In terms of T, how long does it take for the object to accelerate from rest to a speed v now?
Physics
1 answer:
adell [148]3 years ago
5 0

Answer:

It takes \frac{1}{2}T to accelerate the object from rest to the speed v.

Explanation:

From Newton's second law:

F=m\cdot a  (1)

and the definition of acceleration,

\displaystyle{a = \frac{\Delta v}{\Delta t}} (2)

we can solve this problem. Putting (2) in (1) we have:

\displaystyle{F = m\cdot \frac{\Delta v}{\Delta t}} and solving for \Delta t and considering the initial time as zero (t_0=0) and the initial velocity also zero (v_0=0) we have:

\displaystyle{T=\frac{mv}{F}}

Now, for a mass m^*= 2m and the F^*=4F we can wrtie the same equation:

\displaystyle{T^*=\frac{m^*v}{F^*}} and substituting m^* and F^*:

\displaystyle{T^*=\frac{2m\cdotv}{4F}=\frac{2}{4}T=\boxed{\frac{1}{2}T}}

So now, it only takes half the time to accelerate the object from rest to the speed v

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In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast en
g100num [7]

Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω =  22.1 rev/minutes = 22.1 \frac{2\pi }{60} = 2.31 rad/s

mass of rider, m = 54 kg

the speed of a person "stuck" to the wall

v = ω R

  = 2.31 x 6.4

  = 14.8 m/s

the normal force of the wall on a rider

F = m a

a  = ω^2 R

   =  \frac{v^{2} }{R^{2} } R

   = \frac{v^{2} }{R}

F = \frac{mv^{2} }{R}

  = \frac{(54)(14.8)^{2} }{6.4}

  = 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W =  μ N

m g =  μ \frac{mv^{2} }{R}

g = μ \frac{v^{2} }{R}

μ = \frac{gR}{v^{2} }

  = \frac{(9.8) (6.4)}{14.8^{2} }

  = 0.29

5 0
3 years ago
Starting from rest, an intern pushes a 42-kggurney 12 m down the hall with a constant force of 80 N directed downward at an angl
DerKrebs [107]

Answer:

A. The work done by the intern is 792 J.

B. The velocity of the gurney when it has moved 12 m is 6.1 m/s.

C. The 12-m journey takes 3.8 s.

Explanation:

Hi there!

Please see the attached figure for a description of the situation.

<u>Part A: </u>

Work is done by a force when it is applied in the direction of the displacement or against it. In this case, the only force applied in the direction of displacement is the horizontal component of the force applied by the intern.

By trigonometry, the horizontal component of the force is calculated as follows:

cos θ = adjacent/hypotenuse

Looking at the figure, you can notice that the applied force, F, is the hypotenuse of a right triangle and the horizontal component, Fx, is the adjacent side:

cos θ = Fx / F  

Fx = F · cos θ

Fx = 80 N · cos 35°

Fx = 66 N

Now we can calculate the work (W) done by this force:

W = Fx · x

Where x is the displacement:

W = 66 N · 12 m = 792 J

The work done by the intern is 792 J.

<u>Part B:</u>

Applying the work-energy theorem, the work done is equal to the change in kinetic energy:

W = final kinetic energy - initial kinetic energy

Since the gurney starts from rest, the initial kinetic energy is zero. Then:

W = final kinetic energy

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the gurney.

v = velocity.

KE = 792 J

792 J = 1/2 · 42 kg · v²

v²= 2· 792 J / 42 kg

v = 6.1 m/s

The velocity of the gurney when it has moved 12 m is 6.1 m/s.

<u>Part C:</u>

First, let´s find the acceleration of the gurney:

Fx = m · a

Fx/m = a

66N / 42 kg = a

a = 1.6 m/s²

Now using the equation of velocity, let´s find the time at which the gurney has a velocity of 6.1 m/s:

v = v0 + a · t

Where:

v = velocity at time t.

v0 = initial velocity.

a = accleration.

t = time.

v = v0 + a · t

6.1 m/s = 0 + 1.6 m/s² · t

t = 6.1 m/s / 1.6 m/s²

t = 3.8 s

The 12-m journey takes 3.8 s.

3 0
3 years ago
1. A dog of mass 5 kg jumps on a table of mass 30 kg. As the dog walks around on the table, what is the average force that the t
Oksana_A [137]

Answer:

50N

The table is managing to keep the dog up, the dog is not moving up or down so the force that the table applies to the dog must be equal to the dogs weight - I draw this conclusion from Newton's 3rd law.

W= mg

W = 5 × 9.8

W = 49N

round to.one significant figure = 50N

5 0
3 years ago
Help please!!! 100 points
NeX [460]
1. Acceleration
2. If you are going 12 meters per second and you want to know how many seconds it takes yo get to 105 meters, you need yo find how long it takes to go horizontally then vertically. 12 seconds.
3. B
8 0
3 years ago
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Potential energy of an apple is 6j. the apple is 3.00m high. what is the mass of the apple?​
HACTEHA [7]

Answer:

0.2 kg

Explanation:

PE = mgh

6 J = m (9.8 m/s²) (3.00 m)

m = 0.2 kg

5 0
3 years ago
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