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2 years ago

11

a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up

solutions

1 answer:

kykrilka [37]2 years ago

5 0

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

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3 years ago

Can you think of a scenario when the kinetic and gravitational potential energy could both be zero ? Describe or draw how this c

Inga [223]

Both kinetic and gravitational potential energy can become zero at infinite distance from the Earth.

Consider an object of mass <em>m </em>projected from the surface of the Earth with a velocity <em>v. </em>

The total energy of the body on the surface of the Earth is the sum of its kinetic energy $\frac{1}{2} mv^2$ and gravitational potential energy $-\frac{GMm}{R^2}$ .

here, <em>M</em> is the mass of the Earth, <em>R</em> is the radius of Earth and <em>G</em> is the universal gravitational constant.

The gravitational potential energy of the object is negative since it is in an attractive field, which is the gravitational field of the Earth.

The energy of the object on the surface of the earth is given by,

$E_i=\frac{1}{2} mv^2-\frac{GMm}{R^2}$

As the object rises upwards, it experiences deceleration due to the gravitational force of the Earth. Its velocity decreases and hence its kinetic energy decreases.

The decrease in kinetic energy is manifested as an equal increase in potential energy. The potential energy becomes less and less negative as more and more kinetic energy is converted into potential energy.

At a height <em>h</em> from the surface of the Earth, the energy of the object is given by,

$E_h=\frac{1}{2} mv_h^2-\frac{GMm}{(R+h)^2}$

The velocity $v_h$ is less than <em>v</em>.

When h =∞, the gravitational potential energy increases from a negative value to zero.

If the velocity of projection is adjusted in such a manner that the velocity decreases to zero at infinite distance from the earth, the object's kinetic energy also becomes equal to zero.

Thus, it is possible for both kinetic and potential energies to be zero at infinite distance from the Earth. In this case, kinetic energy decreases from a positive value to zero and the gravitational potential energy increases from a negative value to zero.

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3 years ago

1. Sang Hee rubs a balloon on her hair until, when she holds the balloon several inches from

vodomira [7]

The reason why Kim's hair rises and sticks out is due to electrostatic attraction.

<h3>What is charging by friction?</h3>

We know that one of the ways in which a body is able to acquire static charges is by friction. When a body is rubbed against another, there could be loss or gain of charges leaving a net charge on each body.

The process that occurs when some of Kim's hair rises and sticks out toward the balloon, even though the balloon hasn't touched her hair is electrostatic attraction.

Learn more about charging by friction:brainly.com/question/9201910

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1 year ago

regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

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3 years ago

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be $\sqrt{10^3}}$ years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

$1year * \frac{360degrees}{31.62years} = 11.38 degrees$

or about 12 degrees.

6 0

3 years ago