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4 years ago

11

If a rock were dropped from a building it would hit the ground with a certain kinetic energy and velocity. If it fell from a roo

f 25 times as tall, how would its speed of impact compare? You must show work to receive credit.

1 answer:

STALIN [3.7K]4 years ago

5 0

Answer:

no

Explanation:

because if you test it they hit the ground at the same time.

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with what speed does a freely falling object dropped from a height of 88.2m hit the ground? how long does it take for it hit the

maria [59]

Answer:

Vf = 41.6 [m/s]

Explanation:

To solve this problem we must use the equations of kinematics.

Vf² = Vo² + (2*g*y)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

g = gravity acceleration = 9.81 [m/s²]

y = height = 88.2 [m]

Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.

Vf² = Vo² + (2*g*y)

Vf² = 0 + (2*9.81*88.2)

Vf = (1730.48)^0.5

Vf = 41.6 [m/s]

6 0

3 years ago

If earth had no atmosphere , would a falling object ever reach terminal velocity?

Bas_tet [7]

The right answer is no.

6 0

3 years ago

A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During

Artemon [7]

Answer:

Induced EMF, $\epsilon=0.0143\ volts$

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, $B_i=200\ mT=0.2\ T$

Final magnetic field, $B_f=300\ mT=0.3\ T$

The expression for the induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}$

$\epsilon=0.0143\ volts$

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

5 0

3 years ago

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

7 0

4 years ago

Read 2 more answers

A fixed 14.1-cm-diameter wire coil is perpendicular to a magnetic field 0.52 T pointing up. In 0.28 s , the field is changed to

ELEN [110]

Answer:

the average induced emf in the coil is 0.016 V.

Explanation:

Given;

diameter of the wire, d = 14.1 cm = 0.141 m

change in magnetic filed strength, dB = 0.52 T - 0.23 T = 0.29 T

change in time, dt = 0.28 s

The area of the wire is calculated as follows;

$A = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times (0.141)^2}{4} \\\\A = 0.0156 \ m ^2$

The induced emf is calculated as follows;

$emf = \frac{dBA}{dt} \\\\emf = \frac{0.29 \times 0.0156}{0.28} \\\\emf = 0.016 \ V$

Therefore, the average induced emf in the coil is 0.016 V.

8 0

3 years ago