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3 years ago

A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?

2 answers:

8 0

A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps

Scrat [10]3 years ago

6 0

Answer:

The gravitational potential energy is 314 J.

(A) is correct option.

Explanation:

Given that,

Mass of dog = 4 kg

Height = 8 m

We need to calculate the potential energy

Using formula of potential energy

$P.E = mgh$

m = mass

g = acceleration due to gravity

h = height

Put the value into the formula

$P.E=4\times9.8\times8$

$P.E=313.6=314\ J$

Hence, The gravitational potential energy is 314 J.

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A boy is holding a ball 1 m from the ground with a force of 20 N. He holds it still for 60seconds. How much power in watts is be

Sergio [31]

Answer:

Power = 0.33 Watts

Explanation:

Given the following data;

Distance = 1m

Force = 20N

First of all, we would solve for the work done by the boy.

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 20*1 = 20J

Now to find power;

Power = workdone/time

Power = 20/60

Power = 0.33 Watts.

8 0

3 years ago

A mass of 30.0 grams hangs at rest from the lower end of a long vertical spring. You add different amounts of additional mass ΔM

ra1l [238]

Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0 => Fe = m*g

Using Hook's law:

K*X = m*g Solving for K:

K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

K = 10 / 0.0268 = 373.13N/m

3 0

4 years ago

Read 2 more answers

The 10-kg uniform rod is pinned at end

Anton [14]

Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º

3 0

3 years ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago

We can model the human back as a pivoted rod?

3241004551 [841]

Answer:

So the answer is yes, we can the back be shaped like a spinning rod

spinal column that is approximated by a long and narrow rod,

Explanation:

The bone system of the body is very well modeled in physics, the back has a spinal column that is approximated by a long and narrow rod, this rod is fixed in the lower part to the coccyx and has a weight in the upper part (head), this rod has longitudinal vertical movement and twisting movement around the lower part of the bar.

So the answer is yes, we can the back be shaped like a spinning rod

7 0

3 years ago