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VashaNatasha [74]

3 years ago

8

students make a small elevator machine with 5 kg and 10 kg masses on either side how fast will the masses accelerate once they a

re released?

1 answer:

mihalych1998 [28]3 years ago

8 0

Answer:

The acceleration of the elevator machine is, a = 3 m/s²

Explanation:

Given data,

The mass on the one end, m = 5 kg

The mass on the other end, M = 10 kg

According to the Atwood's machine

Ma = Mg - T

ma = T - mg

Adding those equations,

a (M + m) = g ( M - m)

a = (M + m) / ( M - m)

Substituting the values,

a = (10 + 5) / (10 - 5)

= 3 m/s²

The acceleration of the elevator machine is, a = 3 m/s²

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A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago

A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experi

borishaifa [10]

Complete answer

A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100rad/s2. After making 2844 revolutions, its angular speed is 140rad/s

(a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?

Answer:

a) 126.59 radians per second

b) 134.1 seconds

Explanation:

We can use the rotational kinematic equations for constant angular acceleration.

a) For a) let’s use:

$\omega^{2}=\omega_{0}^{2}+2\alpha\varDelta\theta$ (1)

with $\omega_{0}$ the initial angular velocity, $\omega$ the final angular velocity, $\alpha$ the angular acceleration and $\Delta \theta$ the revolutions on radians (2844 revolutions = 17869.38 radians). Solving (1) for initial velocity:

$\sqrt{\omega^{2}-2\alpha\varDelta\theta}=\omega_{0}$

$\omega_{0}^2=\sqrt{(140)^2 -(2)(0.100)(17869.38)=126.59 \frac{rad}{s}}$

b) Knowing those values, we can use now the kinematic equation

$\omega=\omega_{0}+\alpha t$

with t the time, solving for t:

$t=\frac{\omega-\omega_0}{\alpha}=\frac{140-126.59}{0.1}$

$t=134.1 s$

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3 years ago

Best explanation for why oceans have 2 different types of currents<br>

iragen [17]

Answer:

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8 0

3 years ago

Two objects with a Gravitational Force of 5000 N are moved apart 3 times the original distance. What is the new gravitational at

allochka39001 [22]

Gravitational force between two objects will depends on mass and distance between the two objects

It is given by the formula as

$F = \frac{Gm_1m_2}{r^2}$

Now here we can say that it depends inversely on square of distance between two masses

now if force is 5000 N when two masses are at distance "r" from each other

so it is given as

$5000 = \frac{Gm_1m_2}{r^2}$

similarly we can say that if distance is increased by 3 times then we will have

$F = \frac{Gm_1m_2}{(3r)^2}$

now we will find the ratio of two forces

$\frac{F}{5000} = \frac{1}{9}$

so now the new force will be given as

$F = \frac{5000}{9} = 555.5 N$

6 0

3 years ago

consider a charged parallel-plate capacitor. which combination of changes would quadruple its capacitance?

kogti [31]

Its capacitance Double the charge and double the plate area.

Doubling the distance between the plates of a capacitor double the capacitance. Doubling the distance between the plates of a capacitor quadruples the capacitance. As the distance between the plates decreases, the capacitance increases because the potential difference decreases.

Halving the distance between the plates of a parallel plate capacitor doubles the capacitance of the capacitor from its initial capacitance. When two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor with the sum of the plate areas of the individual capacitors. As we saw earlier all other factors being equal, more disk space equals more capacity.

Learn more about Capacitor here:-brainly.com/question/27393410

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4 0

1 year ago