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4 years ago

A child on a swing falls a height of 1m. What speed will the child reach?

Physics

1 answer:

kirill [66]4 years ago

7 0

Answer: $v=4.42\ m/s$

Explanation:

Given

Child falls a height of $h=1\ m$

Applying conservation of energy we can write

Loss in Potential energy=Gain in kinetic energy

$mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 1}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

Thus child reaches a velocity of 4.42 m/s

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What 2 parts of your foot do you use to dribble a soccer ball

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Answer:

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Explanation:

thats how i do it

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3 years ago

A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the sta

Volgvan

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4 years ago

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$