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3 years ago

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three students make the following claims about determining the average speed and average velocity of the object over the entire

time interval depicted in the graph below … which claim is correct?

1 answer:

Mashcka [7]3 years ago

5 0

Answer:

D. none of them po

Explanation:

pl give me points 73

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Pls helppp. Is this right?

andreyandreev [35.5K]

Answer:

yes you are totally right

7 0

3 years ago

Read 2 more answers

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

What is the average velocity of the object from t=1s to t= 3 s?

galina1969 [7]

Answer:

Explanation:

Since this is a distance v time graph, the slope of the line from 1s to 3s is the velocity. However, it looks like, at t=3, the velocity is 0, so getting the definite velocity is not going to happen. We can estimate it as closely as possible. Since the line is tending from the upper left to the lower right, the slope is negative, so the velocity is also negative. That leaves only C or D as our answers. And the slope is closer to -1 than to -5, so choice D. is the one you want.

7 0

3 years ago

The slope of a velocity versus time graph gives

Marina86 [1]

Explanation:

Average of acceleration

4 0

3 years ago

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Virty [35]

Answer:

$\boxed {\boxed {\sf 29,400 \ Joules}}$

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

$E_P= m \times g \times h$

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg
g= 9.8 m/s²
h= 20 m

Substitute the values into the formula.

$E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m$

Multiply the three numbers and their units together.

$E_p=1470 \ kg*m/s^2 \times 20 m$

$E_p=29400 \ kg*m^2/s^2$

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

$E_p= 29,400 \ J$

The crate has <u>29,400 Joules</u> of potential energy.

7 0

3 years ago