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Scilla [17]
3 years ago
11

A skateboarder rolls off a horizontal ledge that is 1.32m high and lands 1.88m from the base of the ledge. What was the initial

velocity? (Unit=m/s)
Physics
1 answer:
NemiM [27]3 years ago
4 0

Answer:

Velocity = 3.622 m/s

Explanation:

Given:

Starting height = 1.32 m

Distance = 1.88 m

Find:

Velocity

Computation:

Using 2nd equation of motion.

s = ut + [1/2]gt²  , where g = 9.8 m/s² and u = 0

1.32 = [1/2][9.8]t²

Time (t) = 0.519 sec

We know that

Velocity = Distance / Time

Velocity = 1.88 / 0.519

Velocity = 3.622 m/s

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An old theory about the creation of the universe has been replaced by a new theory. Which of these statements is correct about t
frez [133]

Answer:

A because it was rejected as it cannot become a law it was totally wrong as compared to today's information about out universe that how big bang theory was there.... So it is right the option A

8 0
1 year ago
Define wheel and axle 4 example of wheel and axle​
natita [175]

Answer:

A system of two co-axial cylindersof different diameters which rotate together is called wheel and axle example; the door knob , knob of the tap ,screw driver,water tap

5 0
3 years ago
Heat is transferred from your body into an ice cube by
Dima020 [189]
2.) <span>Heat is transferred from your body into an ice cube by "Conduction"

3.) </span><span>The loudness of a sound will be determined by the wave’s "Frequency"

4.) </span>The speed of a sound wave <span>"is slower than an electromagnetic wave"

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3 0
3 years ago
Without the wheels, a bicycle frame has a mass of 6.71 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit
dalvyx [7]

Answer:43.311 J

Explanation:

Given

mass of frame\left ( m_f\right )=6.71 kg

mass of  wheel \left ( m_w\right )=0.820 kg

radius of wheel\left ( r\right )=0.343 m

v=3.22 m/s

Moment of inertia of each wheel\left ( I\right )=\frac{1}{2}mr^2[/tex]

I=0.0482 kg-m^2

kinetic Energy of whole cycle=Kinetic energy of wheels and frame+rotational energy of Wheels

K.E.=\frac{1}{2}\left ( m_f+m_w\right )v^2+2\times \frac{1}{2}I\omega ^2

K.E.=\frac{1}{2}\left ( 6.71+0.820\right )3.22^2+2\times \frac{1}{2}0.0482\times \left ( \frac{3.22}{0.343}\right )^2

K.E.=39.037+4.274=43.311J

4 0
3 years ago
Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.3×109 V and the quantity of charg
Georgia [21]

Answer:

A) ΔU = 3.9 × 10^(10) J

B) v = 8420.75 m/s

Explanation:

We are given;

Potential Difference; V = 1.3 × 10^(9) V

Charge; Q = 30 C

A) Formula for change in energy of transferred charge is given as;

ΔU = QV

Plugging in the relevant values gives;

ΔU = 30 × 1.3 × 10^(9)

ΔU = 3.9 × 10^(10) J

B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.

This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.

Thus;

P.E = K.E

ΔU = ½mv²

Where v is final velocity.

Plugging in the relevant values;

3.9 × 10^(10) = ½ × 1100 × v²

v² = [7.8 × 10^(8)]/11

v² = 70909090.9090909

v = √70909090.9090909

v = 8420.75 m/s

4 0
3 years ago
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