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cestrela7 [59]
3 years ago
5

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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A cardboard box sits on top of a concrete sidewalk where the coefficient of friction between the surfaces is 0.4. The mass of th
Sloan [31]

Answer:

Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]

Explanation:

To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.

The gravitational force is equal to:

Fg = (10 * 9.81) = 98.1 [N]

Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.

N - Fg = 0

N = Fg

N = 98.1 [N]

The friction force is defined as the product of normal force by the coefficient of friction.

Ff = N * μ

Ff = 98.1 * 0.4

Ff = 39.24 [N]

By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.

60 - Ff = m * a

60 - 39.24 = 10 * a

a = 2.076[m/^2]

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3 years ago
Which particles do not affect the stability of the atom
lutik1710 [3]
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5 0
3 years ago
In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
A high school bus travels 240 km in 6.0 h. What is its average speed for the trip? (in km/h).
KengaRu [80]

Answer:

40 km/h

Explanation:

First...

Look at the formula speed is equal to the distance over time or s = d/t.

Next...

Use the formula: 240/6.0

Finally...

Solve: 40

So the answer: 40 km/h

5 0
4 years ago
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