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garri49 [273]
4 years ago
11

A stone thrown horizontally from a height of 5.72 m hits the ground at a distance of 13.30 m. Calculate the initial speed of the

stone. Neglect air resistance.
Physics
1 answer:
kakasveta [241]4 years ago
3 0

Answer:

initial velocity=12.31 m/s

Final speed= 16.234 m/s

Explanation:

Given Data

height=5.72 m

distance=13.30 m

To Find

Initial Speed=?

Solution

Use the following equation to determine the time of the stone is falling.  

d = vi ×t   ½ ×9.8 × t²

Where  

d = 5.72m and vi = 0 m/s

so  

5.72 = ½× 9.8 ×t²

t = √(5.72 ÷ 4.9)

t=1.08 seconds

To determine the initial horizontal velocity use the following equation.

d = v×t

13.30 = v ×1.08

v = 13.30 ÷ 1.08

v=12.31 m/s

To determine stone’s final vertical velocity use the following equation

vf = vi+9.8×t............vi=0 m/s

vf = 9.8×1.08

vf= 10.584 m/s

To determine stone’s final speed use the following equation  

Final speed = √[Horizontal velocity²+Final vertical velocity²]

Final speed = √{(12.31 m/s)²+(10.584 m/s)²}

Final speed= 16.234 m/s

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A load of 250 kg is hung by a crane's cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to th
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Answer:

1250\sqrt{3}

Explanation:

Let's look at this in a simple manner, because it is.

The crane weights 250Kg. Okay.

Since it is hung, there is the acceleration of gravity being applied on it (10m/s²)

Since F = m * a

F = 250 * 10

F = 2500

Now we know that the downward Force is 2500N.

To find the force that is being applied on that 30° angle, we can multiply our 2500N by cos30°, which happens to be \frac{\sqrt{3}}{2}.

Therefore, the force pulling the box in the cable's direction is:

\frac{2500\sqrt{3} }{2}  = 1250\sqrt{3}

<em />

<em>If you have any questions feel free to comment.</em>

<em />

<em>Have a great one and mark brainliest if it helped, please</em>

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2 years ago
Two immersion heaters, A and B, are both connected to a 120.0-V supply. Heater A can raise the temperature of 1.00 L of water fr
Inessa05 [86]

Answer:

Ratio of resistance of heater A to resistance of heater B is 5.80

Explanation:

Consider C be the specific heat of water, R₁ and R₂ be the resistance of heater A and heater B respectively.

Given:

Mass of water in heater A, m₁ = 1 L

Mass of water in heater B, m₂ = 5.80 L

Initial temperature, T₀ = 20 ⁰C

Final temperature, T₁ = 90 ⁰C

Time, t = 5 min

Amount of heat required to raise the temperature of water by heaters A and B are given by:

Q₁ = m₁C(T₁ - T₀)       and  

Q₂ = m₂C(T₁ - T₀)

Ratio of power used by both the heaters A and B is:

\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }

Since, time t, temperature difference(T₁ - T₀) and specific heat C are same for both the heaters A and B. So, the above equation becomes:

\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }    ...(1)

The relation to determine electrical power for both heaters A and B are:

P_{1}=\frac{V^{2} }{R_{1} }     and

P_{2}=\frac{V^{2} }{R_{2} }

Here V is the voltage applied to both the heaters and is equal.

So, the ratio of electrical power of heaters is:

\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }     ....(2)

But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,

\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }

Substitute the suitable values in the above equation.

\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }

\frac{R_{1} }{R_{2} }=5.80

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