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2 years ago

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A ball is thrown horizontally from the top of a tower at 30m high, and lands 15m from its base. What is the ball's initial speed

?

1 answer:

PilotLPTM [1.2K]2 years ago

5 0

Answer:

This question requires gravitational acceleration,t=（2h/g）^ 1/2,Level the initial speed is 15/t.

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You are on a sled at the top of a hemispherical, snowy hill of radius 13 m. You begin to slide down the hill. How fast are you m

8_murik_8 [283]

Answer:

Explanation:

There will be loss of potential energy due to loss of height and gain of kinetic energy .

loss of height = R - R cos 14 , R is radius of hemisphere .

R ( 1 - cos 12 )

= 13 ( 1 - .978 )

h = .286 m

loss of potential energy

= mgh

= m x 9.8 x .286

= 2.8 m

gain of kinetic energy

1/2 m v ² = mgh

v² = 2 g h

v² = 2 x 9.8 x 2.8

v = 7.40 m /s

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2 years ago

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What is the net force acting on the piano

CaHeK987 [17]

answer: 500 net force

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3 years ago

When an unbalanced force acts on an object the change in the objects ____or ____ depends on the size and direction of the force

timama [110]

When an unbalanced force acts on an object the change in the object state of rest or motion depends on the size and direction of the force.

If a body is at state of rest or motion, when an unbalanced external force acts on it, its starts moving in the direction of force and magnitude of its velocity or acceleration depends on the magnitude of force applied.

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3 years ago

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A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

2 years ago

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

2 years ago