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antoniya [11.8K]
2 years ago
10

What is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?

Physics
2 answers:
maksim [4K]2 years ago
8 0

Answer:

M=75

Explanation:

33750=1/2x M x30²

30x30=900

1/2x900=450

33750/450=75

nikklg [1K]2 years ago
5 0
The Energy is Kinetic Energy.

Kinetic Energy = 1/2*mv²,  Where m is mass in kg, v is velocity in m/s

Energy is 33750 Juoles,  v = 30m/s

1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²)     Using a calculator

m = 75 kg

Mass of object is 75 kg.
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A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10
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I= \frac{1}{2} c \epsilon_0 E^2
where
I is the intensity
c is the speed of light
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E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C

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(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T

The magnetic force is given by
F=qvB \sin \theta
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(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.

(d) This time, the particle is moving with speed v=3.7 \cdot 10^4 m/s, in a direction perpendicular to the magnetic field (so, the angle \theta is 90^{\circ}), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=
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2 years ago
A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
pochemuha
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Oksana_A [137]

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

Given;

The initial temperature of the coffee T₀ = 95 °C

The temperature of the room = 21°C

Let T be the temperature at time of cooling t in mins

According to Newton's law of cooling;

\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}}  =  \int\limits kdt\\\\Log(T-21) =kt +  Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453

When the temperature is 50 °C, the time t in min is calculated as;

T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins

Therefore, when the temperature of the coffee is 50 °C, the time will be 20.68 mins

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Answer:

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