Which direction do the particles of the medium move as compared to the energy in a transverse wave?

A particular engine has a power output of 8 kW and an efficiency of 37%. If the engine expels 10994 J of thermal energy in each

VARVARA [1.3K]

Answer:

(a) 17450.8 J (B) 1.239 sec

Explanation:

We have given power output =8 KW

Heat expelled = 10994 j

Efficiency =37% = 0.37

(A) We know that efficiency $\eta =1-\frac{heat\ expelled}{heat\ input}$

$0.37=1-\frac{10994}{heat\ input}$

Heat input =17450.8 J

(B) Work done by the engine is = 17450.8-10994=6456.8 j

Power output is given by 8 KW =8000 W

So time for each cycle is $\frac{8000}{6456.8}=1.239 sec$

3 0

3 years ago

A large cube has a mass of 25kg. It is being acclerated

soldier1979 [14.2K]

Answer:

p= 400.29N ........the horizontal force

Explanation:

Given data

mass=25 kg

small cube mass mass=4 kg

Us (The Coefficient of static b/w two cubes) = 0.71

to find

The horizontal force to keep the small cube from sliding downward

Solution

F=ma.........................from Newton Second law

Where F=force

a=acceleration

m=mass

we can write equation in form of acceleration

a=F/m

The acceleration on small box is same as that on the large box.

Let P be force to find.

then:

a=p/(25kg+4kg)

a=p/(29kg)m/s²

The force acting on small box:

F=ma

f=4*(p/29)N........................normal force

friction force= Us*(normal force).........where Us is coefficient of static friction.

friction force= 0.71*(4*p/29)

Now to find weight

weight= mg

weight= 4*9.8

for the object(small box) not to slide down the friction force b/w the two objects have to be exactly the same as the weight of the object.

0.71*(4*p/29)=4*9.8

solving for p(force)

p= 400.29N ........the horizontal force

3 0

3 years ago

Find the reaction supports at Ta and TB as shown in the loaded beam.

koban [17]

ANSWER

$\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}$

EXPLANATION

First, we have to make a sketch of the direction of the moments of the forces about 12m from the left in the diagram:

The sum of upward forces must be equal to the sum of downward forces. This implies that:

$\begin{gathered} T_A+T_B=20+10 \\ T_A+T_B=30N \end{gathered}$

Also, the sum of clockwise moments must be equal to the counter-clockwise moments:

$\begin{gathered} (20\cdot8)+(T_B\cdot4)=(T_A\cdot12) \\ 160+4T_B=12T_A \\ \Rightarrow12T_A-4T_B=160 \end{gathered}$

From the first equation, make TA the subject of the formula:

$T_A=30-T_B$

Substitute that into the second equation:

$\begin{gathered} 12(30-T_B)-4T_B=160 \\ 360-12T_B-4T_B=160 \\ -16T_B=160-360=-200 \\ T_B=\frac{-200}{-16} \\ T_B=12.5N \end{gathered}$

Substitute that into the equation for TA:

$\begin{gathered} T_A=30-12.5 \\ T_A=17.5N \end{gathered}$

Therefore, the reaction supports at TA and TB are:

$\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}$

5 0

1 year ago

An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV.

yuradex [85]

Answer:

3.9E-8

Explanation:

We know that

Mv²/r = Bqv

So

r= mv/Bq

But E is 1/2mv² which is 5.53eV

m²v² =2m x 5.53eV

mv = √( 2 x 9.1E-31)

So

r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19

= 3.9x10-8cm

Explanation:

7 0

3 years ago

Why does a fast working machine possess more power? Clarify your answer<br>Plz help by answering

BARSIC [14]

A chemical reaction increases in speed with the increase in temperature, hard to call that “power” but there's an increase in entropy in the system, far from its equilibrium point molecules start vibrating and colliding more within the system they're in, thus causing the chemical reaction to speed up.
This is a very different answer ....
Hope this helps you....
Mark me as brainliest please...

7 0

3 years ago