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AleksAgata [21]
3 years ago
7

If a force does not act parallel to the resulting displacement, what is the effect on the work done by the force? 

Physics
1 answer:
m_a_m_a [10]3 years ago
8 0
Then only the component of the force that's parallel to the displacement
is used to calculate the work.  The component of force that's perpendicular
to the displacement  doesn't move through any distance at all, so its contribution
to the total work is zero.
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Since rods are about 1000 times more sensitive than cones (at 470 nm), they should be able to detect smaller values of the elect
8090 [49]

Answer:

  a. Rods are about 1000 times more sensitive than cones.

Explanation:

The answer choice shown is a direct copy of the first line of the problem statement. It doesn't appear to be any more complicated than that. (It's a reading comprehension question.)

3 0
2 years ago
A train reaches a speed of 35.0 m/s after accelerating at a rate of 5.00 m/s2 over a distance of 40.0 m. What was the train’s in
MatroZZZ [7]

Answer:

Initial velocity, U = 28.73m/s

Explanation:

Given the following data;

Final velocity, V = 35m/s

Acceleration, a = 5m/s²

Distance, S = 40m

To find the initial velocity (U), we would use the third equation of motion.

V² = U² + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

35² = U + 2*5*40

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U² = 1225 - 400

U² = 825

Taking the square root of both sides, we have;

Initial velocity, U = 28.73m/s

5 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
CAN SOMEONE PLEASE TELL ME WHAT THIS WHEEL IS CALLED.WILL GIVE BRAINLIEST
kolbaska11 [484]

Answer:

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Explanation:

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3 0
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tangare [24]

Answer:

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