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Vlad [161]
3 years ago
10

If a person weighs 90 N on Earth, and the gravitational acceleration due to Jupiter's gravity is 25.98 m/s2, how much would the

person weigh on Jupiter?
A. 3.46 N
B. 238.59 N
C. 882.00 N
D. 2338.20 N
Physics
2 answers:
Elden [556K]3 years ago
8 0

Answer:

A. 3.46

Explanation:

____ [38]3 years ago
7 0

Answer:

B. 238.59 N

Explanation:

First, use their weight on Earth to find their mass:

W = mg

90 N = m (9.8 m/s²)

m = 9.18 kg

Now use this mass to find the new weight on Jupiter.

W = mg

W = (9.18 kg) (25.98 m/s²)

W = 238.59 N

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It takes 20 N of force to move a box a distance of 10 m. How much work is done on the box?
PIT_PIT [208]

Answer:

Ans is 200 J

Explanation:

Given:  Force = 20N

            Distance = 10m

Work done  = Force * displacement

                    =  20 * 10

                    =  200 J

3 0
4 years ago
A 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same cons
LenKa [72]
Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
F=ma
from which we find an expression for the acceleration:
a= \frac{F}{m} (1)

Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
v_f^2 - v_i^2 = 2 a S (2)
where
v_f is the final speed of the object
v_i is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing v_f (since the final velocity of the two objects is zero), we find
-v_i^2 =  2 \frac{F}{m}S
S=- \frac{v_i^2 m}{2F}
where we can ignore the negative sign (because the force F will bring another negative sign).

For the first object, we have
S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} =  \frac{8}{F} [m]
And for the second object we have
S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]

And since the braking force applied to the two objects is the same, the two objects cover the same distance.
3 0
4 years ago
A car is traveling at a speed of 37 m/s.
Vika [28.1K]

1. The speed in kilometers per hour (Km/h) is 133.2 Km/h

2. Yes, the speed is exceeding the 125 Km/h limit

<h3>How to convert 37 m/s to Km/h</h3>

From the question given above, the following data were obtained:

  • Speed (in m/s) = 37 m/s
  • Speed (in Km/h) =?

We can convert 37 m/s to kilometers per hour (Km/h) by doing the following:

1 m/s = 3.6 Km/h

Therefore,

37 m/s = 37 × 3.6

37 m/s = 133.2 Km/h

Thus, 37 m/s is equivalent to 133.2 Km/h

<h3>2. How to determine if the speed is exceeding the limit</h3>
  • Speed of car = 133.2 Km/h
  • Speed limit = 125 Km/h

From the above, we can see that the speed of the car is greater than the speed limit.

Thus, we can conclude that the speed of the car is exceeding the speed limit.

Learn more about conversion:

brainly.com/question/10893215

#SPJ1

3 0
2 years ago
A person of Mass 50kg is on a swing which has the speed 10m,/s at the lowest point of the motion the ropes of the swing are 2.5m
atroni [7]

Answer:

T  =  2490 newton

Explanation:

Given that,

Mass of a person, m = 50 Kg

Speed of the swing at the lowest point, v = 10 m/s

Length of the rope, r = 2.5 m

Tension formula is given by

                            T = mg + ma  N

where,              T - Tension in the string

                         m - mass of the body

                         g - acceleration due to gravity

                          a - acceleration of the body

Since the person is swinging, the acceleration of the body is given by

                          a = v²/r m/s²

                             = 10/2.5  m/s²

                          a = 40 m/s²

Substituting in T

                   T  =  50 Kg x 9.8 m/s²  + 50 Kg X 40 m/s²

                       = 490 + 2000 Kg m/s²

                   T  =  2490 newton

Therefore the tension in the rope is 2490 newton

8 0
4 years ago
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
aliina [53]

Answer:

v_f = 10.38 m / s

Explanation:

For this exercise we can use the relationship between work and kinetic energy

          W = ΔK

note that the two quantities are scalars

Work is defined by the relation

          W = F. Δx

the bold are vectors.  The displacement is

          Δx = r_f -r₀

          Δx = (11.6 i - 2j) - (4.4 i + 5j)

          Δx = (7.2 i - 7 j) m

 

          W = (4 i - 9j). (7.2 i - 7 j)

remember that the dot product

           i.i = j.j = 1

           i.j = 0

           

           W = 4  7.2 + 9  7

           W = 91.8 J

the initial kinetic energy is

           Ko = ½ m vo²

           Ko = ½ 2.0 4.0²

           Ko = 16 J

we substitute in the initial equation

          W = K_f - K₀

          K_f = W + K₀

          ½ m v_f² = W + K₀

          v_f² = 2 / m (W + K₀)

          v_f² = 2/2 (91.8 + 16)

           v_f = √107.8

          v_f = 10.38 m / s

3 0
3 years ago
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