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3 years ago

What is a newton equal to in terms of mass and acceleration

1 answer:

8 0

1 newton is the force that accelerates 1 kilogram of mass at the rate of 1 meter per second^2. / / / 1N = 1 kg-m / sec^2 .

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The flywheel of a motor has a mass of 300.0 kg and a radius of 55 cm. If the motor can exert a torque of 2000.0 N . m on this fl

Alexeev081 [22]

Answer:

Explanation:

Work done by torque is given as

Word one = torque × angular displacement

W = τ × θ

Given that,

τ = 2000Nm

Mass of motor = 300kg

Radius r = 55cm = 0.55m

Work done by wheel in first t= 23second.

Now we need to find the angular displacement

We know that,

τ = I•α

Moment of inertia of wheel

I = MR²

I = 300 × 0.55²

I = 90.75 kgm²

Then, τ = I•α

α = τ / I

α = 2000/90.75

α = 22.04rad/s²

Then, using circular motion,

∆θ = wit + ½αt²

wi = 0rad/s

∆θ = ½αt²

∆θ = ½ × 22.04 × 23²

∆θ = 5829.2 rad.

Then,

Work done?

W = τ × θ

W = 2000 × 5829.2

W = 1.17 × 10 ^7 J

7 0

2 years ago

Which best describes a nonmagnetic material?

Natali [406]

Material that is not attracted to metal

3 0

2 years ago

Read 2 more answers

A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km

Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

$v_f=v_o+at$

Since the car starts at rest, you can write the following equation:

$20=0+a(5) \\\\a=20\div 5=4 m/s^2$

Now that you have the acceleration, you can do this:

$d=v_o+\dfrac{1}{2}at^2$

Once again, there is no initial velocity:

$d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m$

Hope this helps!

8 0

2 years ago

According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind

Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0

2 years ago

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

$x=v\cos\theta t$

On substituting the known values,

$\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}$

The change in the height, y of the basketball is given by,

$y=-v\sin\theta t+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity.

On substituting the known values,

$\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}$

Final answer:

The parametric equations describing the shot are

$\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}$

8 0

8 months ago