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9966 [12]
3 years ago
6

A weightlifter can exert an upward force of 3750 N. If a dumbbell has a mass of 225 kg, what is the maximum number of dumbbells

this weightlifter could hold simultaneously if he were on the Moon? (The Moon's acceleration due to gravity is approximately 0.166 times freefall acceleration on Earth.) The weightlifter cannot pick up a fraction of a dumbbell, so make sure your answer is an integer.
Physics
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

10 dumbbells

Explanation:

First we need to calculate the gravity on the moon.

0.166 x 9.8 m/s² = 1.627 m/s²  (the gravity is 0.166 times the Earth's gravity)

Taking this acceleration due to gravity, and multiplying it by the mass of a single dumbbell.

F = m x a

F = (225 kg) x (1.627 m/s²)

F = 366.075 N  (the amount of force exerted by the dumbbell)

Taking the weightlifter's total upward force and dividing it by the force exerted by one dumbbell, we can calculate the amount of dumbbells that can be carried.

(3750 N) / (366.075 N)

= 10.24 dumbbells  (but since there cannot be a fraction of a dumbbell, the answer is <u>10 dumbbells</u>).

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Option 2nd is the answer
3 0
2 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0
3 years ago
Emmit is lifting a box vertically which forces are necessary for calculating the total force
zavuch27 [327]

Answer:

FG and FP

Explanation:

Gravitational Force(FG) because the box is being pulled down and a resistance force pushing up because he is pushing(FP) up on the box.

I hope this helps!

4 0
3 years ago
A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c
tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

B=\dfrac{\mu_{0}NI}{l}

Where, N = Number of turns

I = current

l = length

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}

B=0.005026=5.026\times10^{-3}\ T

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

4 0
3 years ago
If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:
madam [21]

Answer:

kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

Explanation:

Given the data in the question;

we know that;

Kinetic energy = 1/2.mv²

given that mass of the object is doubled; m1 = 2m

speed is halved; v1 = V/2

Now, New kinetic energy will be; 1/2.m1v1²

we substitute

Kinetic Energy = 1/2 × 2m × (v/2)²

Kinetic Energy = 1/2 × 2m × (v²/4)

Kinetic Energy = 1/2 × m × (v²/2)

Kinetic Energy = 1/2 [ 1/2mv² ]

Kinetic Energy = 1/2 [ KE ]

Therefore; kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

3 0
2 years ago
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