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2 years ago

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Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

it reaches the bottom of the ramp

1 answer:

Arturiano [62]2 years ago

4 0

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s$

So, its velocity at the bottom of the ramp is 2.21 m/s.

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A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pres

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Answer:

C) The pressure reading stays the same.

Explanation:

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Un pendule est constitue par une masse ponctuelle m= 0,1kg accrocher a un fil sans masse de longueur L = 0,4 m on ecarte ce pend

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2 years ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

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2 years ago

The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.

Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

$W=0.24\times 1\times 10^3=240\ W/m^2$

Energy required by the United States is $1\times 10^{19}\ J/yr$ (assumed)

Power

$P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W$

Area

$A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2$

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

$1\ m^2=\frac{1}{1000\times 1000}\ km^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2$

Converting to mi²

$1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2$

Each side of the square would be 1320.33 kilometers or 509.79 miles

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2 years ago

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th

mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

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2 years ago