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a_sh-v [17]
3 years ago
5

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

What would be the weight of a space probe on the surface of Jupiter?
A.2.5 times lighter than on Earth
B.6.25 times heavier than on Earth
C.2.5 times heavier than on Earth
D.6.25 times lighter than on Earth
Physics
1 answer:
Ann [662]3 years ago
6 0

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

W=m\times g

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

W'=mg'=m\times 2.5g

\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}

W'=2.5\times W

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

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A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera
Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m=2.1\times 10^3 kg

Initial velocity,u=0

Distance,s=5.9 m

\theta=19^{\circ}

Average friction force,f=4.0\times 10^3 N

We have to find the speed of the car at the bottom of the driveway.

Net force,F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3

Where g=9.8 m/s^2

Acceleration,a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}

v=\sqrt{2as}

v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}

v=3.9 m/s

7 0
3 years ago
Which should not be a part of scientific inquiry?
lisabon 2012 [21]

I would say;

a - bias.

8 0
3 years ago
Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the
Rzqust [24]

Wow !  This is not simple.  At first, it looks like there's not enough information, because we don't know the mass of the cars.  But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

                                  PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

                                 KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down.  So now, here comes the big jump.  Put a comment under
my answer if you don't see where I got this equation:

                                   KE = 0.9  PE

        (1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)     

Divide each side by (mass): 

               (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass.  As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

               (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

               Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

                          =  (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

                          =  (24,500 / 88.2)  (m²/s²) / (m/s²)

                          =        277-7/9    meters
                                  (about 911 feet)
3 0
3 years ago
`i have a sealed cylinder sitting in my lab that contains 1,000.0 ml of gas. if i compress the cylinder and change the volume to
marin [14]
Because the temperature remains constant, we can apply Boyle's Law which states that 
pV = constant
where
p = pressure
V = volume

Define the two states of the gas.

State 1
Pressure = p₁
Volume = 1000 ml

State 2
Pressure = p₂
Volume = 500 ml

Apply Boyle's law.
1000p₁ = 500p₂
2 = p₂/p₁

By halving the volume, the pressure doubles.

Answer:
The pressure increases by a factor of 2.

7 0
3 years ago
When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the
dolphi86 [110]

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

5 0
3 years ago
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