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IRISSAK [1]
3 years ago
7

A 0.68 kg squirrel is resting on a branch 8 meters above the ground. What is the gravitational potential energy of a squirrel? A

: 2.27 J. B :5.44 J. C :21.76 J. D :53.312 J
Physics
1 answer:
ANEK [815]3 years ago
7 0

Answer:

The gravitational potential energy of a squirrel is 53.312 J.

Explanation:

We have,

Mass of a squirrel is 0.68 kg

It is placed at a height of 8 m above the ground.

It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

E=mgh\\\\E=0.68\times 9.8\times 8\\\\E=53.312\ J

So, the gravitational potential energy of a squirrel is 53.312 J.

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Match Newton's laws with their descriptions. 1. Every applied force is opposed by an equal force. 2 .A force must be applied to
Marina CMI [18]
A would be number 2. Newton's First Law states that an object at rest, will stay at rest and an object in motion, will stay in motion, unless acted upon by an unbalanced force. B would be number 3. His Second Law states that <span>the sum of the forces acting on a body is equal to the product of the mass of the body and the acceleration produced by the forces. And, C would be number 1. His Third Law states that for every action, there is an equal and opposite reaction. Hope this helps!</span>
6 0
4 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c
Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

4 0
3 years ago
When you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Ea
Klio2033 [76]

Answer:

5.66 × 10⁻²³ m/s

Explanation:

If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.

Since initial momentum = final momentum,

mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.

My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?

So mv₁ + M × 0 = m × 0 + MV₂

mv₁ = MV₂

V₂ = mv₁/M =  54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s

5 0
3 years ago
How does the law of conservation of mass apply to this reaction: Al + 3HCl → H2 + AlCl3 ?
11Alexandr11 [23.1K]

Answer:

Only the hydrogen needs to be balanced. There are equal numbers of aluminum and chlorine.

Explanation:

I took the K12 test, listened to some other idiot and got it wrong, so this is the correct one.

Hope this helps!

8 0
3 years ago
Read 2 more answers
Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place
Semmy [17]

Answer:

Ans= 9

See attached picture for clearer solution.

Explanation:

The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .  

The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .  

The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .  

The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .  

The ratio of the largest to the smallest net force = 9/4*2/2  /  1/4 2/2 . = 9

5 0
3 years ago
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