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3 years ago

A truck pushes a mound of dirt 5 meters with a force of 75 newtons. How much work has been done ?

Physics

1 answer:

Paladinen [302]3 years ago

6 0

Work = force × distance, assuming that the force is parallel to displacement.

Work L=75×5=375J

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What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300

Vinvika [58]

The efficiency of an ideal Carnot heat engine can be written as:
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the temperature of the cold region
$T_{hot}$ is the temperature of the hot region

For the engine in our problem, we have $T_{cold}=300 K$ and $T_{hot}=400 K$ , so the efficiency is
$\eta= 1 - \frac{300 K}{400 K}=0.25$

4 0

3 years ago

The angle of incidence always equals what?

Ratling [72]

Answer:

Angle Of Incidence Formula The angle of incidence is equal to the reflected angle through the law of reflection. The angle of incidence and the angle of reflection is always equal, and they are both on the same plane along with the normal. 2,44,451

Explanation:

that could help you with work.

4 0

3 years ago

Read 2 more answers

You place a solid cylinder of mass M on a ramp that is inclined at an angle β to the horizontal. The coefficient of static frict

zlopas [31]

Answer:

Explanation:

Let the frictional force required be f .

frictional force is responsible in creating rotational motion in the cylinder.

torque created by frictional force = f R

if angular acceleration be α

α = f R / I , I is moment of inertia of cylinder .

α = a / R , a is linear acceleration.

f R / I = a / R

a = f R² / I

linear acceleration a of cylinder down the slope

ma = mgsinθ - f , ( f force is acting upwards and mgsinθ is acting downwards )

mf R² / I = mgsinθ -f

f ( m R² / I + 1) = mgsinθ

f = mgsinθ / ( m R² / I + 1)

= mgsinθ / ( m R² / mk² + 1) , k is radius of gyration of cylinder.

= mgsinθ / ( R² / k² + 1)

Putting the given values

f = Mgsinβ /( R² / k² + 1)

for cylinder , R² / k² = 2

f = Mgsinβ /3

6 0

3 years ago

A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ri

Karo-lina-s [1.5K]

Answer:

Explanation:

Given

Mass of mother $m=79\ kg$

Mother is released with a speed of $v=2.5\ m/s$

Assuming we need to find elastic Potential energy stored in Trampoline

Kinetic energy acquired by Mother comes from Elastic Potential Energy stored in the spring

I.e. Elastic Potential Energy=Kinetic Energy of mother

$\frac{1}{2}kx^2=\frac{1}{2}mv^2$

$E=\frac{1}{2}\times 79\times 2.5^2=246.875\ J$

So 246.875 J of Energy is stored in the trampoline as he pull the ring

5 0

3 years ago

HELP FAST

katen-ka-za [31]

Answer:

C. 2 and 4

Explanation:

my teacher went over it and the answer was that

3 0

2 years ago