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goblinko [34]
3 years ago
6

What is the full form of work?​

Physics
1 answer:
Pepsi [2]3 years ago
7 0

Answer: JOB

Explanation: They full forms of work are JOB which is JUST OBEY BOSS or Joining Other Bussiness .

You might be interested in
A sprinter speeds up to 3 m/s during the last 2 seconds of the race with an
quester [9]

Answer:

<em>The initial speed of the sprinter was 2.2 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

The following relation applies:

v_f=v_o+at

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The sprinter speeds up from an unknown initial speed to vf=3 m/s in t=2 seconds with an acceleration of a=0.4~m/s^2.

To find the initial speed, we solve the equation for vo:

v_o=v_f-at

Substituting the values:

v_o=3-0.4*2

v_o=3-0.8

v_o=2.2~m/s

The initial speed of the sprinter was 2.2 m/s

4 0
3 years ago
Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The
Mnenie [13.5K]

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round  Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

Answer: So, the final angular momentum is 27.73 revolution per minute

7 0
3 years ago
Write a message about an important aspect of life on earth to be sent beyond our solar system
zepelin [54]
We grow crops (plants) on earth, so it's very important for that to go beyond earth because if there is no way to grow crops, we won't get enough nutrients from some synthetic (fake) food. hope I helped! :)
6 0
3 years ago
A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?
alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y=0 m is the final height of the ball

y_{o}=60 m is the initial height of the ball

V_{o}=0 m/s is the initial velocity (the ball was dropped)

g=9.8 m/s^{2} is the acceleratio due gravity

t is the time

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (2)

t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}} (3)

Finally we find the time the ball is in the air:

t=3.49 s (4)

7 0
3 years ago
4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer
Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

3 0
3 years ago
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