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3 years ago

11

A construction worker pushes a cement block with a 750 N force for 10 minutes without the cement block moving. How much work did

the construction worker do during the 10 minutes

2 answers:

lutik1710 [3]3 years ago

8 0

The work done by a system on a different body is equal to the product of the force exerted and the distance that the body has move in parallel to the force exerted. In this item, we have to determine first the distance and multiply it with the given force equal to 750N.

luda_lava [24]3 years ago

8 0

Answer:

Zero

Explanation:

Work done is defined as the product of force applied and the displacement traveled by the object.

Work = force applied in the direction of displacement x displacement

The SI unit of work is Joule. Work may be positive, negative or zero.

here, force applied = 750 N, displacement = 0 m

So, work done = 750 x 0 = 0 Joule

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You are out running and the first mile takes you 10 minutes. The second mile takes you 20 minutes. This is an example of

cupoosta [38]

If Equal distance is traveled in equal interval of time then it is known as uniform motion in which velocity of object will remain same.

Then if distance covered will be same and the time taken to cover same distance is decreasing then it shows that speed is increasing with time due to which it took less time to cover same distance. This is also known as positive acceleration.

Now if the distance covered will be same and time taken to cover same distance is increasing then it shows that speed is decreasing with time due to which it took more time to cover the same distance. This is also known as negative acceleration.

Now in the above case it is given that the first mile takes you 10 minutes. The second mile takes you 20 minutes. So the time taken is increasing while we cover same distance so this is an example of <u>Negative Acceleration</u>

5 0

3 years ago

The air in a car tire la compressed when the car rolls over a rock. If the air

stealth61 [152]

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

8 0

3 years ago

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

2 years ago

57. Estimate Potential Energy A boulder with a

Oksanka [162]

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

3 0

3 years ago

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

$T^2=\frac{4\pi ^2}{GM} a^2$

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

$M = 3.5M_{sun} = 6.96*10^{30} kg$

$a=4.2AU = 6.28*10^{11} meters$

$G = 6.67*10^{-11}m^3kg^{-1}s^{-2}$

We put this into Kepler's law and get:

$T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.$

This when converted to years is 4.6 years.

4 1

3 years ago

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