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Ronch [10]
3 years ago
13

PLEASE HELP

Physics
1 answer:
Mashutka [201]3 years ago
4 0

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

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How can a calculated height be greater than an actual height?
sammy [17]

Answer:

mesuring heigh and weight is important

3 0
2 years ago
Uncle Harry weighs 750 N. What is his mass in kg?
Elenna [48]

Answer:

W = MG

750 = M * 10

M = 750/10

M = 75 kg

8 0
3 years ago
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What is difference between non uniform and uniform circular motion?
statuscvo [17]

Answer:

The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.

Explanation:

The speed of the object traveling in uniform motion is constant, the actual speed and the average speed of the moving body is same.

6 0
3 years ago
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0
3 years ago
Does parallax affect the precision of a measurement that you make
Paul [167]
Yes, parallax affects the precision of a measurement that you make. It introduces an error in the order of the parallax. It will cause the measurement to be different from the real answer. Hope this answers the question. Have a nice day.
8 0
3 years ago
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