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2 years ago

Suppose kinetic energy is 4 joules of a moving object what will happen if we increase the speed twice

Physics

1 answer:

Artyom0805 [142]2 years ago

3 0

Answer:

The kinetic energy is 16 joules.

Explanation:

Kinetic energy (E) is given by:

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

Where:

m: is the mass

v: is the speed

When the kinetic energy is 4 joules, the speed of the object is:

$v_{1}^{2} = \frac{2E_{1}}{m} = \frac{8}{m}$

Now, if the speed is increased twice then we have:

$E_{2} = \frac{1}{2}mv_{2}^{2} = \frac{1}{2}m(2v_{1})^{2} = \frac{1}{2}m[4(\frac{8}{m})] = 16 J$

Therefore, the kinetic energy is 4 times the initial value.

I hope it helps you!

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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

nata0808 [166]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

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3 years ago

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 8.00 ✕ 106 m/s pe

KonstantinChe [14]

Answer:

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Explanation:

It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

$\frac{m_ev_e^2}{r}=qv_eB\\\frac{m_pv_p^2}{r}=qv_pB\\$

Take the ratio:

$m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_e}{m_p}v_e\\\Rightarrow v_p=\frac{1}{1840}\times 8.0\times 10^6 m/s\\\Rightarrow v_p=4347.8m/s$

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