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3 years ago

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

Physics

1 answer:

pav-90 [236]3 years ago

5 0

<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

Displacement, s = 0 m

Acceleration, a = -9.81 m/s²

Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

s = ut + 0.5 at²

0 = 128.89 x t + 0.5 x (-9.81) x t²

t² - 26.28 t = 0

t ( t- 26.28) = 0

t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

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nikitadnepr [17]

- The integers 1,2,3,4,5,6,7,8,9 are always significant

- Zeros in between numbers are always significant

Explanation:

The rules to determine whether a digit is significant or not are the following:

- All the non-zero digits are significant: 1,2,3,4,5,6,7,8,9

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2 years ago

A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of

nlexa [21]

Answer:

The given phenomenon supports the Principle of Conservation of momentum.

Explanation:

law of conservation of momentum

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum = m1*u1 + m2*u2

and we are given

m1 = mass of car = 600 kg

m2 = mass of van = 400 kg

u1 = Initial Speed of Car

to find the initial speed we use equation of motion

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum = 7200 Ns .........(1)

final momentum = m₁v₁ + m₂v₂

where,

v₁ = v₂ = final speed of car+van (both locked ) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns -------- (2)

on comparing (1) and (2)

Initial momentum = Final Momentum

Hence, the phenomenon of the system supports the principle of conservation of momentum.

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3 years ago

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2 years ago

A 5 Kg bowling ball is thrown at a stationary 1.6 Kg bowling pin at 5 m/s. If the final velocity of the ball is 2.5 m/s. The fin

Sergio039 [100]

Answer: hope this helps

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2 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

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3 years ago