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zzz [600]
2 years ago
9

You weigh 580 N on Earth. If you were to go to Mars, where its gravitational pull is 3 . 7 11 m /s 2 , what would you weigh? (Hi

nt: gravity on Earth is 9.8 m /s 2 .
Physics
2 answers:
andrew-mc [135]2 years ago
7 0

Answer:

59.18 kg

Explanation:

use f=ma

f= 580 N

a = 9.8 m/s 2

weigh(m) doesn't change only force(F) changes

AnnZ [28]2 years ago
4 0

Answer: 219.6N

Explanation:

Find the mass of this person by dividing his weight by the accelleration. Remember, F=ma

580 = m*9.8

m = 59.18kg

Then multiply the mass by the new gravitational acceleration on Mars.

59.18*3.711 = 219.6N

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Answer:

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A.) X: Load, Y: Fulcrum, Z: Lever

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3 years ago
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Maggie completed a 10000-m race at an average speed of 160
Gala2k [10]

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

3 0
3 years ago
James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than
Molodets [167]

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h =  c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

7 0
3 years ago
5. If one object has a greater speed than a second object. does the first necessarily have a greater acceleration? Explain, usin
Sveta_85 [38]

Answer:

Explanation:

5. not necessarily so that the first object could have left with initial velocity and the second not, so even if the second has a greater acceleration its velocity is less than that of the first

6. The acceleration of the motorcycle is

     SI System Reductions

     Vo = 80 km / h (1000m / 1km) (1h / 3600s) = 22.2 m / s

     Vf = 90 km / h (1000m / 1km) (1h / 3600s) = 25 m / s

     Vf = Vo + at at = Vf-Vo

     am = (Vf-Vo) / t

     am = (25 -22.2) / t = 2.8 / t

      am= 2.8/t

For the bike we have

      Vf = 10 km / h (1000m / 1km) (1h / 3600s) = 2.78 m / s

      Vo = 0

      ab = (Vf -Vo) / t

      ab = (2.78 -0) / t

      ab = 2.8/t

Since time is the same for both of us, if we round to Significant figures the two accelerations are equal

7. If when an object is slowing or slowing down.

     For example, a car goes north and must stop at the traffic light, the acceleration of the brakes goes south

8. Yes, since an object can go to the left and the acceleration to the right, but the object will lose speed over time

9. in the launch of projectiles the acceleration is negative and the speed after half the path is also negative

10. Car B must be moving to car A, because if they leave together B has more acceleration, bone that travels the distance at the same time

11. When we have friction, the velocity of an object increases by an external force, but the friction also increases the acceleration, but since it is positive, the velocity increases until the acceleration is zero and hence the velocity remains constant.

8 0
3 years ago
In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir
Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be S_{1}

and the distance covered by Sir Alfred be S_{2}

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

S = ut + \frac{1}{2}at^{2}

Where S is the distance traveled

u is the initial velocity

a is the acceleration

and t is the time

For Sir George,

S = S_{1}

u = 0 m/s (Since they start from rest)

a =0.21 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{1}  = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1}  = 0.105 t^{2}\\

t^{2} = \frac{S_{1}}{0.105}

Now, for Sir Alfred

S = S_{2}

u = 0 m/s (Since they start from rest)

a =0.26 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{2}  = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2}  = 0.13 t^{2}\\

t^{2} = \frac{S_{2}}{0.13}

Since, they traveled for the same time, t just before collision, we can write

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

S_{1} + S_{2} = 86 m

Then, S_{2} = 86 - S_{1}

Then,

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13} becomes

\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}

0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}

S_{1} = 38.43 m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0
3 years ago
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