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Firdavs [7]
3 years ago
15

HI CAN ANYONE PLS ANSWER DIS PLS!!!!!!

Physics
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

Volume of an object (L): blue 105.00, yellow 105.00, green 102.50, red 101.25

Density of an object (kg/L) Mass / Volume: blue 1, yellow 1, green 1.014, red .992

Sink or Float: blue sink, yellow sink, green float, red float

hope this helps

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Which of the following is an example of Newton’s second law of motion?
inysia [295]

Answer:

B

Explanation:

Newton’s Second Law of Motion

Newton’s Second Law of Motion states that ‘when an object is acted on by an outside force, the mass of the object equals the strength of the force times the resulting acceleration’.

This can be demonstrated dropping a rock or and tissue at the same time from a ladder. They fall at an equal rate—their acceleration is constant due to the force of gravity acting on them.

The rock's impact will be a much greater force when it hits the ground, because of its greater mass. If you drop the two objects into a dish of water, you can see how different the force of impact for each object was, based on the splash made in the water by each one.

5 0
3 years ago
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What evidence indicates that a chemical change took place when the iron and sulfur combined to form iron sulfide? A. The element
fgiga [73]

Answer:

Explanation:burn the gas

5 0
3 years ago
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How do you rationalize the tension being used in Tennis Racket strings using the concept of impulse and momentum?
zheka24 [161]

Answer:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F_{axial}

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

3 0
3 years ago
A block of mass 0.84 kg is suspended by a string which is wrapped so that it is at a radius of 0.061 m from the center of a pull
lora16 [44]

Answer:

E_l = 1.713 J

Explanation:

Given data:

mass of block is M_b = 0.84 kg

radius of block = 0.061 m

moment of inertia is 6.20 \times 10^{-3} kg m^2

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum  we have

M_b g D = \frac{1}{2} M_b v^2 + \frac{1}{2} I \omega^2 +  E_{loss}

0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times  0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l

solving for energy loss

E_l = 1.713 J

3 0
3 years ago
A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3s later. How high
Marysya12 [62]

We use kinematic equation,

h=ut+\frac{1}{2} gt^2

Here, h is vertical height, u is initial vertical velocity, t is time taken and g is acceleration due to gravity.

As diver dives out horizontally, his velocity is directed horizontally; that is, the initial vertical velocity is 0. So above equation becomes

h=\frac{1}{2} gt^2

Given, t =3 s.

Therefore,

h=\frac{1}{2}\times 9.8m/s^2(3\ s)^2 =0.5\times 9.8 m/s^2\times 9 s^2\\\\h=44.1\ m.

Now the horizontal distance of the diver to hit the water from base,  

=1.8\ m/s \times 3\ s=5.4\ m

6 0
4 years ago
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