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2 years ago

What would the velocity of an object be if it was on a 1.75m

Physics

1 answer:

JulijaS [17]2 years ago

4 0

Answer:

T = 4 sec / 2 = 2 sec period of revolution

S = 2 pi R = 2 * pi * 1.75 m = 11 m

V = S / T = 11 m / 2 sec = 5.5 m/s speed of object

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A tennis ball (m=0.060 kg) is moving horizontally at 20 m/s toward a tennis player who hits it straight back at 26 m/s. What is

Rus_ich [418]

Answer:

0.36 kg-m/s

Explanation:

Given that,

Mass of a ball, m = 0.06 kg

Initial velocity of the ball, u = 20 m/s

Final velocity of the ball, v = 26 m/s

We need to find the change in momentum of the tennis ball. It is equal to the final momentum minus initial momentum

$\Delta p=m(v-u)\\\\=0.06\times (26-20)\\\\=0.36\ kg-m/s$

So, the change in momentum of the ball is 0.36 kg-m/s.

4 0

2 years ago

7. If the magnitude of the gravitational force of

JulijaS [17]

Newton's Third Law states that for every action there is an opposite and equal reaction:

If the gravitational force of the Earth on the Moon is F then the gravitational force of the Moon on the Earth is also F

3 0

3 years ago

Photoelectrons with a maximum speed of 8.00 • 106 m/sec are ejected froma surface in the presence of light with a frequency of 6

Andru [333]

The kinetic energy is given by:

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

We know the mass and the maximum speed, plugging their values in the expression above we have:

$\begin{gathered} K=\frac{1}{2}(9.1\times10^{-31})(8\times10^6)^2 \\ K=2.91\times10^{-17}\text{ J} \end{gathered}$

Therefore, the answer is d.

5 0

1 year ago

An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to

mafiozo [28]

Answer:

wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

w_f = w₀ - a t

w₀ = w_f + a t

w₀ = 109 + 4.7 1.87

w₀ = 117.8 rad / s

let's reduce to revolutions / s

w₀ = 117.8 rad / s (1 rev / 2pi rad)

w₀ = 18.75 rev / s

8 0

3 years ago

What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0

3 years ago