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finlep [7]
3 years ago
6

you get a flat tire on her car while driving. You use a jack to change the tire. It exerts a force of 5,000 N to lift the car 0.

25 m. How much work is done by the jack?
Physics
1 answer:
ch4aika [34]3 years ago
6 0

Answer:

b is the anwser

Explanation:

you are super

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Before Mt. Everest was discovered, what was the highest mountain in the world?
Juli2301 [7.4K]
Kangchenjunga (8,586 metres (28,169 ft)) was considered to be the highest mountain from 1838 until 1852. Mount Everest, 8,848 metres (29,029 ft). Established as highest in 1852 and officially confirmed in 1856.
3 0
3 years ago
A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

6 0
3 years ago
A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?
Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

W=(mg)h

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

W=(1250)(2)=2500 J

Now we can calculate the power, which is equal to the work done per unit time:

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

P=\frac{2500}{3}=833.3 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

5 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
Harrizon [31]

1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().

The Rydberg formula is used to determine the energy change.

Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.

aaΔE=R(1n2f−1n2i) aa

were

2.17810-18lJ is the Rydberg constant.

The initial and ultimate energy levels are ni and nf.

As a change of pace from

n=5 to n=3 gives us

ΔE

=2.178×10-18lJ (132−152)

=2.178×10-18lJ (19−125)

=2.178×10-18lJ×25 - 9/25×9

=2.178×10-18lJ×16/225

=1.549×10-19lJ

Learn more about Rydberg formula here-

brainly.com/question/13185515

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8 0
2 years ago
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