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Vaselesa [24]
3 years ago
10

Jupiter is a much more massive planet than Earth. What would happen to a person’s mass and weight if he were on Jupiter?

Physics
2 answers:
Ksenya-84 [330]3 years ago
7 0

The mass would not change because your mass is how much space you take up not how much you weigh.

Harrizon [31]3 years ago
5 0
There would be no mass or weight and he would float away
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Ne4ueva [31]

Answer:

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3 0
2 years ago
While playing a scavenger hunt game, Anthony walks 48.0 m to the south and then walks 12.0 m to the west. What single straight-l
Alenkasestr [34]
As these are distances created by moving in a straight line, using a trigonometric analysis can solve the missing single straight-line displacement. Looking at the 48m and 12m movements as legs of a triangle, obtaining the hypotenuse using the pythagorean theorem will yield us the correct answer.
This is shown below:

c^2 = 48^2 + 12^2
c = sqrt(2304 + 144)
c = sqrt(2448)
c = 49.48 m

To obtain the angle at which Anthony walks 49.48, we obtain the arc tangent of (12/48). This is shown below:

arc tan (12/48) =14.04 degrees.

Therefore, Anthony could have walked 49.48 m towards the S 14.04 W direction.
4 0
3 years ago
How does a physicist answer a scientific question?
lara31 [8.8K]

Answer:

b

Explanation:

the answer is B

3 0
2 years ago
A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into
vagabundo [1.1K]

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2}  + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s

5 0
3 years ago
Read 2 more answers
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
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