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siniylev [52]
2 years ago
11

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. if the ball hits the ground 4.0 second

s later, approximately how high is the cliff?
Physics
1 answer:
Viktor [21]2 years ago
8 0

Answer:

78.4 m

Explanation:

To obtain the height of the cliff;

We can use the Relation to obtain the final velocity, v

v = u + at

a = acceleration due to gravity = 9.8m/s²

v = 0 + (9.8*4)

v = 0 + 39.2

v = 39.2 m/s

To obtain the Height, S

v² = u² + 2aS

39.2^2 = 0 + 2(9.8)S

39.2^2 = 0 + 19.6S

1536.64 = 19.6S

S = 1536.64 / 19.6

S = 78.4 m

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Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000°C. To remove this en
Alona [7]

Answer:

thats a lot, which one u want me to do?

Explanation:

8 0
2 years ago
10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends
MaRussiya [10]

Explanation:

(10) Mass of a soccer player, m = 0.42 kg

Initial speed, u = 0

Final speed, v = 32.5 m/s

Time, t = 0.21 s

We need to find the force that sends soccer ball towards the goal.

Force, F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.42 \times (32.5-0)}{0.21}\\\\F=65\ N

So, 65 N of force soccer ball sends towards the goal.

(11) Mass of the satellite, m = 72,000 kg

Initial speed, u = 0 m/s

Final speed, v = 0.63 m/s

Time, t = 1296 s

We need to find the force is exerted by the rocket on the satellite.

Force, F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{72,000\times (0.63-0)}{1296}\\\\F=35\ N

So, 35 N of the force is exerted by the rocket on the satellite.

Hence, this is the required solution.

3 0
2 years ago
The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro
ASHA 777 [7]

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

4 0
2 years ago
An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal
alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   V = \epsilon - Ir

now,

   IR = \epsilon - Ir

   I(R+r) = \epsilon

   I= \dfrac{\epsilon}{R+r}

inserting the values

   I= \dfrac{12}{0.05+0.01}

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 I= \dfrac{\epsilon}{R+r}

   I= \dfrac{12}{0.14+0.01}

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W

6 0
3 years ago
Read 2 more answers
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
Burka [1]

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

v(4) = -15.7\,\frac{m}{s}

8 0
3 years ago
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