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gizmo_the_mogwai [7]
3 years ago
9

A spring-loaded toy gun projects a 1305 g nerf pellet horizontally. The spring constant is 8.8 N/m, the barrel of the gun is 17

cm long, and a constant frictional force of 0.026 N exists between the barrel and the nerf pellet. If the spring is compressed 6.0 cm for this launch, determine the speed of the pellet as it leaves the barrel. (Assume the pellet is in contact with the barrel for the full length of the barrel.)
Physics
1 answer:
Marina86 [1]3 years ago
8 0

Answer:

v_f=2.15x10^{-4}m/s

Explanation:

Given:

m=1305g*\frac{1kg}{1000g}=1.305kg

K=8.8N/m

L=17cm*\frac{1m}{100cm}=0.17m

x_i=6.0cm*\frac{1m}{100cm}

ΔE=Wf

E_f-E_i=W_f

\frac{1}{2}*m*v^2-\frac{1}{2}*K*x_i^2=-2*F_k*L

m*v_f^2=-2*F*L+K*x_i^2

Solve to vf'

v_f=\sqrt{\frac{-2*F_k*L+K*x_i}{m}}

v_f=\sqrt{\frac{-2*0.026N*0.17m8.8N/m*(0.06m)^2}{1.305kg}}

v_f=2.15x10^{-4}m/s

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A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Nutka1998 [239]

Answer:

6.99535\times 10^{-6}\ V/m

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r = Radius = 35000000 m

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

Intensity of Electric radiation is given by

I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2

Intensity of Electric radiation is given by

I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m

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4 years ago
An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
Wewaii [24]

Answer:

\frac{dB}{dt} = 3.03 \times 10^6 T/s

Explanation:

As we know that the power emitted by the source is given as

P = 170 W

now we know that

P = \frac{N}{t} (\frac{hc}{\lambda})

now we know that energy density is given as

u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}

now we have

E = B c

u = \frac{B^2}{2\mu_0}

intensity is defined as

I = \frac{P}{A}

now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

here we have

I = \frac{P}{4\pi r^2}

I = \frac{170}{4\pi (410)^2}

I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

\frac{dB}{dt} = 3.03 \times 10^6 T/s

4 0
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