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tester [92]
3 years ago
14

The car A has a weight of 4000 lb and is traveling to the right at 3 ft/s. Meanwhile a 2000 lb car B is traveling at 6 ft/s to t

he left. If the cars crash head-on, and at a time instant during the crash the impact force on A is 900 lb to the left, what is the magnitude and direction of the impact force exerted on B at the instant?
(A) 900 lb to the left
(B) 450 lb to the left
(C) 450 lb to the right
(D) 1800 lb to the left
(E) 900 lb to the right
Physics
1 answer:
iVinArrow [24]3 years ago
6 0

To solve this problem we should apply Newton's third law for which it is defined that there must be an equal reaction in the opposite direction.

From this law, if Car A generates a force on car B, that car must have opposed a force exactly the same but in the opposite direction. The car A moved to the left and generates a force of 900lb so the magnitude of the force in the car B is also 900lb but to the right (opposite direction to the first car)

The correct option is 900lb to the right.

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Using Planck’s constant as h = 6.63 E-34 J*s, what is the wavelength of a proton with a speed of 5.00 E6 m/s? The mass of a prot
Marrrta [24]
De Broglie's identity gives the relationship between the momentum and the wavelength of a particle:
p=mv= \frac{h}{\lambda}
where
p is the particle momentum
m is its mass
v its velocity
h is the Planck constant
\lambda is the wavelength

By re-arranging the equation, we get
\lambda=  \frac{h}{mv}
and by using the data about the proton, given in the text, we can find the proton's wavelength:
\lambda= \frac{h}{mv} = \frac{6.63 \cdot 10^{-34} Js}{(1.66 \cdot 10^{-27} kg)(5.00 \cdot 10^6 m/s)} =7.99 \cdot 10^{-14} m
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3 years ago
If you were to be drawn into a black hole, what would happen? To the black hole, not to you.
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Answer:

It grows

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The blacks holes will absorb

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3 years ago
A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The
IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0
3 years ago
A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram
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The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

\rm F=qvBsin\theta

where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

F=6.6\times 10^{-3}\ N

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

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