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3 years ago

A sled is at rest at the top of a slope 2 m high. The sled has a mass of 45 kg. What is the sled's potential energy?

Physics

1 answer:

kap26 [50]3 years ago

8 0

Answer:

PE = 882 J

Explanation:

Through the International System we know that the gravity of the earth is 9.8 m/s², so...

Data:

m = 45 kg
g = 9.8 m/s²
h = 2 m
PE = ?

Formula:

$\boxed{\bold{PE=m*g*h}}$

Replace and solve:

$\boxed{\bold{PE=45\ kg*\ 9.8\frac{m}{s^{2}}*2\ m}}$

$\boxed{\boxed{\bold{PE=882\ J}}}$

The potential energy of the sled is <u>882 Joules</u>.

Greetings.

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kramer

Answer:

60.2 J

Explanation:

Efficiency is the ratio of work out to work in.

e = Wout / Win

0.86 = Wout / 70 J

Wout = 60.2 J

5 0

3 years ago

Read 2 more answers

Calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the mean radius of the earth at

valentinak56 [21]

Answer: $M = 5.98\times 10^{24} kg$

Explanation:

We know that force acting on an object due to Earth's gravity on the surface is given by:

$mg = G\frac{Mm}{r^2}\\ \Rightarrow g = \frac{GM}{r^2}$

where g is the acceleration due to gravity, r would be radius of Earth, M is the mass of Earth and G is the gravitational constant.

It is given that at pole, g = 9.830 m/s² and r = 6371 km = 6371 × 10³ m

$\Rightarrow M = \frac{g\times r^2}{G}$

$M = \frac{9.830 m/s^2 \times (6371 \times 10^3 m )^2}{6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}}$

$\Rightarrow M = 5.98\times 10^{24} kg$

Hence, Earth's mass is $5.98\times 10^{24} kg$

5 0

3 years ago

A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte

zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

5 0

3 years ago

A mass of gas under constant Pressure occupies a volume of 0.5 M3 At a temperature of 20ﾟC using the formula for Cubic expansion

MAVERICK [17]

To answer this question, we must use the equation for the volumetric expansion of gases at constant pressure. This equation is given by:

$V=V_{0}(1 + \alpha(T_{2}-T_{1}))$

We know:

$V_{0}$ is the initial volume $= 0.5 m ^ 3$

ΔT is the temperature change = 45 ° -20 ° = 25 °

$\alpha$ is the coefficient of gas expansion and is equal to 1/273

Then the final volume of the gas is:

$V=0.5*(1+\frac{1}{273}*25)$

$V=0.546m^ 3$

7 0

3 years ago

An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t

lana [24]

Answer:

Equal to the weight of the object

Explanation:

There are two forces acting on the object hanging on the string:

- The tension in the string, T, acting upward

- The weight of the object, mg, acting downward

We can write Newton's second law as:

$T-mg=ma$

where a is the acceleration of the object.

Here we are told that the elevator is moving at constant speed: this means that the acceleration of the object is zero, therefore a = 0 and the equation becomes

$T=mg$

therefore, the tension in the string is equal to the weight of the object.

7 0

3 years ago