PHYSICS HIGH SCHOOL

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

Answers

Answer 1

Answer:

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

$W=m* g$

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

Weight of the space probe on earth = W'

$W'=mg'=m* 2.5g$

$(W')/(W)=(m* 2.5g)/(m* g)$

$W'=2.5* W$

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.

Answer 2

Answer:

Answer:

2.5 times heavier than on Earth

Explanation:

Related Questions

If a photon has a frequency of 1.15 x 10 to the 15 power Hertz what is the energy of the photon given: Planck's constant is 6.63 x 10 to the negative 34 joule seconds
Vitellium (Vi) has the following composition:Vi-188: 187.9122 amu; 10.861%Vi-191: 190.9047 amu; 12.428%Vi-193: 192.8938 amu; 76.711%Based on this data, what can you predict about the average atomic mass of vitellium?
Which type of electromagnetic radiation cannot be focused? A. Gamma rays B. X-rays C. Ultraviolet D. Infrared
Which of the following statements about geothermal energy is not true?a. It is not a practical source of energy. b. It is an important source of energy in volcanically active areas. c. It works best when the magma pool is near the surface of Earth. d. It is a renewable source of energy.
Anxiety disorders are

You kick a ball with a speed of 14 m/s at an angle of 51 degrees. How far away does the ball land?A. 21.3 m
B. 19.6 m
C. 1.8 m
D. 3.96 m

Answers

Answer: The ball will land at 19.6 m

Explanation:

To calculate the horizontal range of the ball, we use the formula:

$R=(v^2\sin 2\theta)/(g)$

where,

v = velocity of the ball = 14 m/s

$\theta$ = angle at which the ball is thrown = 51°

g = acceleration due to gravity = $9.8m/s^2$

Putting values in above equation, we get:

$R=((14)^2* \sin (2* 51))/(9.8)\n\nR=19.6m$

Hence, the ball will land at 19.6 m

d may be the correct answer because speed is very low so ball should cover less distance but angle is not very high so it can cover little much distance

A girl that has a mass of 30 kg is hanging from a bar motionless. Determine the following forces. If no force exists, enter 0. (use 10m/s2 for gravity) Weight Newtons
Applied force Newtons
Frictional Force Newtons
Normal Force Newtons
Tension Force
Net Force Newtons

Answers

Final answer:

The weight of the girl is 300 N. The applied force, frictional force, and normal force are all zero. The tension force in the bar is 300 N, and the net force is zero.

Explanation:

The weight of the girl can be calculated using the formula:

Weight = mass × gravity

where mass = 30 kg and gravity = 10 m/s2. Therefore, Weight = 30 kg × 10 m/s2 = 300 N.

Since the girl is motionless, the net force acting on her is zero. In this case, the tension force in the bar equals the weight of the girl, which is 300 N. The other forces (applied force, frictional force, and normal force) are also zero since the girl is not moving.

Therefore, the forces are:

Weight: 300 N
Applied force: 0 N
Frictional force: 0 N
Normal force: 0 N
Tension force: 300 N
Net force: 0 N

Learn more about forces here:

brainly.com/question/33301608

#SPJ1

Your friend won the race that you were both running. Instead of being gracious, your friend made fun of you for losing, and now you don't want to run anymore. What is this an example of? Positive peer pressure Negative peer pressure Positive peer influence Negative peer influence

Answers

Answer: Option (d) is the correct answer.

Explanation:

When my friend made fun of me for loosing the race then obviously this thing left me with no confidence. Also, I would feel shame for losing.

As a result, i have decided not to run anymore.

This represents an example of negative peer influence as the comments or fun made by my friend has made me come to a negative decision.

negative peer influence

A proton moves along the x axis according to the equation x = 38 t + 14 t2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous acceleration of the proton at t = 3.0 s.

Answers

Answer:

a) 240 m

b) 122 m/s

c) 28 m/s²

Explanation:

Given:

Equation for motion

x = 38t + 14t²

a) average velocity during first 3 seconds

average velocity = $\frac{\textup{change in displacement}}{\textup{cahnge in time}}$

now,

distance, at t = 0 s

x = 38 × 0 + 14 × 0² = 0 m

distance, at t = 3 s

x = 38 × 3 + 14 × 3² = 240 m

therefore,

average velocity = $(240-0)/(3-0)$ = 80 m/s

b) instantaneous velocity of the proton at t = 3.0 s

Instantaneous velocity, v = $(dx)/(dt)=38+28* t$

Instantaneous velocity, v = $(dx)/(dt)=38+28* 3$

= 122 m/s

c) instantaneous acceleration of the proton at t = 3.0 s

Now,

Acceleration = $(dv)/(dt)$ = 0 + 28 = 28 m/s²

You throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the ground 115 m from the base of the building? How fast must you have thrown the rock?
How high up must the 7th floor be?
If you had thrown it at the same speed but at the 28th floor, how far from the base of the building would it land?

Answers

1) 31.1 m/s

The rock has been thrown straight out of the window: its motion on the horizontal direction is simply a uniform motion, with constant speed $v$ , because no forces act in the horizontal direction. The speed in a uniform motion is given by

$v=(S)/(t)$

where S is the distance traveled and t the time taken.

In this case, the distance by the rock before hitting the ground is $S=115 m$ and the time taken is $t=3.7 s$ , so the initial speed is given by

$v=(115 m)/(3.7 s)=31.1 m/s$

2) 67.1 m

In this part of the problem we are only interested in the vertical motion of the rock. The vertical motion is a uniformly accelerated motion, with constant acceleration $a=9.8 m/s^2$ (acceleration of gravity) towards the ground. In a uniformly accelerated motion, the distance traveled by the object is given by

$S=(1)/(2)at^2$

where t is the time. Substituting a=9.8 m/s^2 and t=3.7 s, we can find S, the vertical distance covered by the rock, which corresponds to the height of the 7th floor:

$S=(1)/(2)(9.8 m/s^2)(3.7 s)^2=67.1 m$

3) 230.1 m

The height of the 7th floor is 67.1 m. So we can assume that the height of each floor is

$h=(67.1 m)/(7)=9.6 m$

And so, the height of the 28th floor is

$h=28\cdot 9.6 m=268.8 m$

We can find the total time of the fall in this case by using the same formula of the previous part:

$S=(1)/(2)at^2$

In this case, S=268.8 m, so we can re-arrange the formula to find t

$t=\sqrt{(2S)/(g)}=\sqrt{(2(268.8 m))/(9.8 m/s^2)}=7.4 s$

And now we can consider the motion of the rock on the horizontal direction: we know that the rock travels at a constant speed of v=31.1 m/s, so the distance traveled is

$S=vt=(31.1 m/s)(7.4 s)=230.1 m$