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3 years ago

What's a sentence using work in a nonscientific sense

Physics

2 answers:

kvv77 [185]3 years ago

7 0

My Physics teacher keeps giving us way too much work to do at home.

serg [7]3 years ago

4 0

Work, like calories can give a person energy. The glophcus which comes from the Greek word for camel which is located in the brain uses the work to make enegy.

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He graph below shows the velocity f(t) of a runner during a certain time interval:

Annette [7]

<span>The following that describes the intercepts on the graph is "The initial velocity of the runner was 4 m/s, and the runner stopped after 8 seconds." It is because the starting point of the line is at 4 and then the ending point is at 8.

</span>

5 0

2 years ago

Read 2 more answers

Give two examples of vernier calliper

omeli [17]

Answer:

Example 1, if a vernier caliper output a measurement reading of 2.13 cm, this means that: The main scale contributes the main number(s) and one decimal place to the reading

E.g. 2. 1 cm, whereby 2 is the main number and 0.1 is the one decimal place number

Explanation:

plz mark as brainliest and hope it helps you

4 0

2 years ago

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

$y(t)=2.80t+0.61t^3$

The velocity of the crate vs time is given by the derivative of the position, so it is:

$v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2$

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

$a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t$ [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

$F(t)=ma(t)$

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

$a(4.10)=3.66\cdot 4.10 =15.0 m/s^2$

And therefore, the force on the crate is:

$F=ma=(5.00)(15.0)=75 N$

7 0

3 years ago

A stunt-rider on a motorcycle rides down a ramp and into a vertical loop-the-loop. If the diameter of the loop is 7.50 m, then t

VikaD [51]

Answer:

v = 6.06 m/s

Explanation:

In order for the rider to pass the top of the loop without falling, his weight must be equal to the centripetal force:

$Centripetal Force = Weight \\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v = \sqrt{gr}$

where,

v = minimum speed of motorcycle at top of the loop = ?

g = acceleration due to gravity = 9.8 m/s²

r = radius of the loop = diameter/2 = 7.5 m/2 = 3.75 m

Therefore, using these values in equation, we get:

$v = \sqrt{(9.8\ m/s^2)(3.75\ m)}$

5 0

3 years ago

During the final stages of descent, asky diver with an

OLEGan [10]

Answer:

Gravity: downwards

Air drag and air-pressure on the inner surface of the the parachute: Upwards

Explanation:

If a sky-diver is in the final stages of his descend with open parachute such that the wind is calm and it does not blows him laterally.

In such a condition the air resistance in the form of drag and the pressure force due to the air captured in the parachute are acting in the upward direction which balance the force of gravity on the body. But this situation may occur momentarily and then again the diver must begin to slowly descend.

3 0

3 years ago