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3 years ago

An object is 45.0 kg here on Earth. What is its mass on Jupiter if the acceleration

Physics

1 answer:

puteri [66]3 years ago

8 0

Answer:

B. 45.0 kg

Explanation:

On earth the object has ;

Mass = 45 kg

acceleration = 9.81 m/s²

On Jupiter, acceleration is 24.79 m/s²

The mass of this object on Jupiter will be 45 kg. It will not change. Mass of an object will only change when you remove part of the object from it or add to it another part. The mass is the same on Earth and on Jupiter. However, due to increased acceleration on Jupiter , the weight will change/ increase because;

Weight = mass * acceleration

<u>On Earth </u>

Weight of the object will be : 45 * 9.81 = 441.45 kg

<u>On Jupiter</u>

Weight of the object will be : 45*24.79 =1115.55 kg

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An athlete wants to refill an old punching bag used for training, as shown in the picture. When refilled, the bag will rip if it

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The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states action and reaction are equal and opposite. That is the force applied to an object is equal in magnitude to force experienced by the object but in opposite direction.

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Learn more about Newton's third law of motion here: brainly.com/question/25998091

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2 years ago

gold has a density of 19.32g/cm^3. aluminum has a density of 2.7g/cm^3. which is heavier: 1g of gold or 1 g of aluminum? explain

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3 years ago

Steam enters an adiabatic turbine at 6 MPa, 600 ℃, and 80 m/s and leaves at 50 kPa, 100 ℃, and 140 m/s. If the power output of t

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Answer:

W(r,out) = 5.81 MW

$\eta$ = 86.1 %

Explanation:

we use here steam table for get value of h1, s1 etc

so use for 6MPa and 600 degree

Enthalphy of steam h1 = 3658.8 kJ/kg

Entropy of steam s 1 is = 7.1693 kJ /kg.K

and

for 50 kPa and 100 degree

Enthalphy of steam h2 = 2682.4 kJ/kg

Entropy of steam s2 is = 7.6953 kJ /kg.K

so we use here energy balance equation that is

$m\times(h1 + \frac{v1^2}{2} = m\times(h2 + \frac{v2^2}{2} + W(out)$ ..............1

put here value and we get m

m = $\frac{5\times1000}{3658.8-2682.4+\frac{80^2-140^2}{2}\times \frac{1}{1000}}$

solve it we get

m = 5.156 kg/s

so by energy balance equation

m $\psi$ 1 = m $\psi$ 2 + W(r,out)

W(r,out) = m( $\psi$ 1 - $\psi$ 2)

W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)

W(r,out) = $m[h1-h2+ \frac{v1^2-v^2}{2}- To (s1-s2)$

W(r,out) = W(a,out) - m.To.(s1-s2) ........................2

put here value

W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)

W(r,out) = 5908.19 = 5.81 MW

and

second law deficiency is

$\eta$ = $\frac{W(a,out)}{W(r,out)}$ ..............................3

put here value

$\eta$ = \frac{5}{5.81}

$\eta$ = 86.1 %

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3 years ago