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2 years ago

Which objects have both potential and kinetic energy? Check all that apply.

Physics

2 answers:

wariber [46]2 years ago

3 0

A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.

denis23 [38]2 years ago

3 0

Answer:

a sky diver who is midway through his descent to the ground

a roller coaster car halfway down the second hill

a falling raindrop

Explanation:

From the law of conservation of energy, it can neither be created nor destroyed but can be converted from one form to another.

The sum of kinetic energy and potential energy is mechanical energy. A body has potential energy due to its position or configuration and kinetic energy due to its speed.

A body can have both potential energy and kinetic energy.

The correct options are:

a sky diver who is midway through his descent to the ground
a roller coaster car halfway down the second hill
a falling raindrop

A boulder resting on top of a mountain has just potential energy.

A parked car also has just potential energy.

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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

Mice21 [21]

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

2 years ago

A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an

inna [77]

Answer:

It will take 4 sec rock to comes its original point

Explanation:

It is given that the rock comes to its original point

So displacement S = 0 m

Initial velocity u = 19.6 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

According to second equation of motion $h=ut+\frac{1}{2}gt^2$

$0=19.6\times t+\frac{1}{2}\times 9.8t^2$

$19.6=4.9t$

t = 4 sec

3 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

2 years ago

A string, 68 cm long and fixed at both ends, vibrates with a standing wave that has 3 antinodes. the frequency is 180 hz. what i

REY [17]

The given situation below describes a standing wave because the string is fixed at both ends. A standing wave having three anti-nodes will have a wavelength that is two-thirds the length of the string. After getting the wavelength, this can be multiplied with the frequency to get the wave speed.

For this problem:

wave length = (2/3)(length of string: 68 cm)
wave length = (10/3 cm)

wave speed = wave length x frequency
wave speed = (10/3 cm) x (180 Hz)
wave speed = 600 cm/s or 0.6 m/s

7 0

2 years ago