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timurjin [86]
3 years ago
12

Caleb is investigating the effect of friction on the motion of an object. He uses the following supplies for the investigation:

Physics
1 answer:
Annette [7]3 years ago
7 0
Part a
Place the ramps on the floor and time each ramp for time it takes for wooden block too reach thee floor. The times will be different and thus you can conclude that friction will increase the time
Part b
Caleb uses the same block as this is his control variable that he wants to keep same so that the results are correct and accurate
Part c
The independent variable is the thing you change before you test so it will be the surface on which the wooden block is sliding
The dependent variable is the variable you measure so that will be the time it takes for the block to slide to the bottom.
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At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops
jasenka [17]

Answer:

force = 11.33 kg-m/s^{2}

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

P_{1}=mU

P_{1}= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion

F=\frac{\Delta P}{\Delta t}

F = \frac{P_{1}-P_{2}}{T}

Kgm/s^2

F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}[/tex]

8 0
4 years ago
An electron enters a region of space containing a uniform 2.71 × 10 − 5 2.71×10−5 T magnetic field. Its speed is 197 197 m/s and
Andrews [41]

Answer:

r = 0.0414mm

F = 757,692.3Hertz

Explanation:

If the body enters space with uniform magnetic field B, the force experienced by the object is expressed as

F = qvBsintheta... 1

Also, if the body undergoes a circular motion, the force experienced by the body in a circular path is given as

Fc = mv²/r... 2

Equating both forces

F = Fc

qvBsin theta = mv²/r

Since the body enters perpendicular to the field, theta = 90°

The equality becomes;

qvB sin90° = mv²/r

qvB = mv²/r

qB = mv/r

r = mv/qB

Given mass of the electron m = 9.11×10^-31kg

Velocity of the object v = 197m/s

Charge on the electron q = 1.6×10^-19C

Magnetic field B = 2.71×10^-5T

Substituting this value into the equation to get the radius r we have;

r = 9.11×10^-31 × 197/1.6×10^-19 × 2.71×10^-5

r = 1794.67×19^-31/4.336×10^-24

r = 413.89×10^-7

r = 0.0000414m

r = 0.0414mm

b) Frequency of the motion F = w/2π where w is the angular velocity

Since w = v/r

F = (v/r)/2π

F = v/2πr

F = 197/2π(0.0000414)

F = 757,692.3Hertz

6 0
4 years ago
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