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noname [10]
3 years ago
5

a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to

bring him/her to a stop​
Physics
1 answer:
Natalka [10]3 years ago
4 0

Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

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In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
hammer [34]

Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

7 0
3 years ago
In this experiment you will investigate which of the following properties of Faraday's law of electromagnetic induction? (Select
11Alexandr11 [23.1K]

Answer:

Answered

Explanation:

Part A

According to Faraday's law the induced emf in coil is equal to negative of its rate of change of magnetic flux time the number of turns in the coil.

\epsilon = -N\frac{d\phi}{dt}= -N\Delta\frac{BA}{\Delta t}

When an emf generated by a change of magnetic flux, produced current of whose magnetic field opposes the change  which produces it.

By the above equation the correct options are 1,2 and 4

Part B

Large signals of  frequency of 60Hz are measured by osciloscope.

Hence the correct option is part 1.

3 0
3 years ago
when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results
Nikitich [7]

When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

  • Hard stabilization is the prevention of erosion through the use of artificial barriers.
  • Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
  • Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
  • These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
  • Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.

Learn more about the Hard stabilization with the help of the given link:

brainly.com/question/16022736

#SPJ4

4 0
2 years ago
What is An object that changes position?
mel-nik [20]

Explanation:

Motion is when an object changes position over time. The object in motion is usually in front of a reference point-an object that appears to stay in one place. The rate at which an object moves is called speed. Speed depends on both time and distance. The velocity of an object is how fast it is going in one direction

How do you know if an object has changed position?

changes position requires a point of reference. An object changes position if it moves relative to a reference point. To visualize this, picture yourself competing in a 100-m dash. You begin just behind the start line

6 0
3 years ago
A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d
Lilit [14]

Answer:

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut + \frac{1}{2}a_yt²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850  + \frac{1}{2}×a_y×(0.850)²

1.29 = 0.36125a_y

a_y = 1.29 / 0.36125

a_y = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a_y

where m₁ is the mass of the text book ( 2.10 kg )

a_y is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N

3 0
3 years ago
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