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2 years ago

Can you answer this math homework? Please!

Physics

1 answer:

notka56 [123]2 years ago

5 0

$\large \mathfrak{Solution : }$

Average speed :

$\boxed{ \dfrac{distance \: \: covered}{total \: \: time \: \: taken} }$

$\dfrac{100}{8}$

$12.5 \: \: m/s$

The horizontal (flat) line on a Distance - time graph shows that the body is at rest because the value of Distance (on y - axis ) is same for increasing time ( on x - axis ), it basically shows that the object is at the same position with the change in time.

therefore, correct option is :

$a. \: \: the \: \: object \: \: is \: \: not \: \: moving$

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

Which events may occur when ocean salinity increases? Check all that apply.

Pavlova-9 [17]

The answer is number two, number four, and number one

3 0

2 years ago

Read 2 more answers

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0

3 years ago

What are some solutions to global temperature changes?<br>help asap

Nimfa-mama [501]

Answer:

Power your home with renewable energy.

Weatherize, weatherize, weatherize.

Invest in energy-efficient appliances.

Reduce water waste.

Actually eat the food you buy—and make less of it meat.

Explanation:

6 0

3 years ago