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3 years ago

5

-How can a camera help someone see something he or she wouldn't normally be able to see?

1 answer:

joja [24]3 years ago

4 0

Answer:

》How can a camera help someone see something he or she wouldn't normally be able to see?

- Although the human eye is able to observe fast events as they happen, it is not able to focus on a single point of time. We cannot freeze motion with our eyes. With a camera, however, so long as there is enough light, we can freeze motion. ... The camera can capture 'the moment', while your eye cannot.

》In what way is laser different from light produced by other sources?

-The major difference between laser light and light generated by white light sources (such as a light bulb) is that laser light is monochromatic, directional and coherent. ... Light from a conventional sources, such as a light bulb diverges, spreading in all directions.

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Calculate the time it would take a motor with a power of 0.6kW to lift a suitcase of 20kg a distance of 50cm

True [87]

Answer:

0.163 s

Explanation:

Appying,

P = mgh/t................ Equation 1

Where P = power of the motor, m = mass of the suitcase, h = vertical distance, t = time, g = acceleration due to gravity.

make t the subject of the equation,

t = mgh/P................ Equation 2

Given: m = 20 kg, h = 50 cm = 0.5 m, P = 0.6 kW = 600 W

Constant: g = 9.8 m/s²

Substitute these values into equation 2

t = (20×0.5×9.8)/600

t = 0.163 s

7 0

3 years ago

If the forces acting on an object are balanced, the object may change speed, change direction, or do both. Please select the bes

Leviafan [203]

False is ur answer sorry if i am too late

8 0

4 years ago

Read 2 more answers

A horizontal force F~ is applied to a block of mass m = 1 kg placed on an inclined

vova2212 [387]

Hi there!

To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.

Recall:

$\large\boxed{F_D = \mu N}}$

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

$\large\boxed{N = Mgcos\theta + Fsin\theta}}$

We can plug in the known values to solve for one part of the normal force:

N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F

Now, we can plug this into the equation for the dynamic friction force:

Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F

For a block to move with constant speed, the summation of forces must be equivalent to 0 N.

If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:

Fcosθ = 1.697 + .1F

Solve for F:

Fcos(30) - .1F = 1.697

F(cos(30) - .1) = 1.697

F = 2.216 N

6 0

3 years ago

Can someone please help me

siniylev [52]

Answer:

M a = (M1 + M2) a = F Newton's Second Law

F = (M2 - M1) g net force on the system

a = (M2 - M1) / (M1 + M2) g

a = (9 - 7) / (9 + 7) g = 2 / 16 * 10.0 m/s^2 = 1.25 m/s^2

6 0

2 years ago

A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

sergiy2304 [10]

Answer:

Acceleration of the car will be $a=0.1375m/sec^2$

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car $F_R=7340N$

So net force on the car $F_{net}=F-F_R=7560-7340=220N$

According to second law of motion we know that F=ma

So $220=1600\times a$

$a=0.1375m/sec^2$

7 0

3 years ago