A box with a mass of 12.5 kg sits on the floor. how high would you need to lift it for it to have a gpe of 355 j

jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a

Morgarella [4.7K]

Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

   = (11 kg) · (3.5 m/s)² / (0.6 m)

   = (11 kg) · (12.25 m²/s²) / (0.6 m)

   = (11 · 12.25) / 0.6 kg-m/s²

   = 224.58 newtons. (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by. It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second. If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.

6 0

2 years ago

Read 2 more answers

Why are force fields used to describe magnetic force?

Illusion [34]

Answer:

If I'm not working I think the answer is C.

7 0

2 years ago

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

2 years ago

1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

azamat

Answer:

$\dot{W} = 339.84 W$

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

$\dot{W} =\dot{m}\frac{V^{2}}{2}$

here $\dot{m}$ is mass flow rate and given as

$\dot{m} = \rho*Q$

$\dot{W} =\rho*Q\frac{V^{2}}{2}$

Putting all value to get minimum power

$\dot{W} =1.18*9*\frac{8^{2}}{2}$

$\dot{W} = 339.84 W$

7 0

2 years ago

A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he

Iteru [2.4K]

Answer:

30.93 m/s

Explanation:

Given that, the speed of stolen car is,

$v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s$

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, $v_{p}$

The policeman catches the stolen car at a distance of,

$S=60km\\S=60000m$

Now the distance covered by the policeman in time t is $v_{p}\times t$

And the distane cover by the thief in stolen car in time(t+60s) is $v_{s}\times (t+60sec)$ .

And these distances are equal and they are equal to 60000 m.

Therefore,

$v_{p}\times t=v_{s}\times (t+60sec)=60000m$

Therfore,

$v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s$

Now use this value to solve for minimum speed of policeman's car.

$v_{p}\times 1940=60000\\v_{p}=30.93 m/s$

Therefore minimum speed of policeman's car is 30.93 m/s.

6 0

2 years ago