What is the change in internal energy if 60J of heat are released from a system and 20J of work is done on the system? Use U=Q-W

Nicolo` works on weekends at a Slow Food Parlor. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6

Slav-nsk [51]

Answer:4.22 J

Explanation:

Given

mass of pitcher $m=2.6\ kg$

Force applied $F=8.8\ N$

distance moved $s=48\ cm$

Applying work-Energy theorem which states that work done by all the forces is equal to the change in kinetic energy of the object

Work done by force $W=F\cdot s$

W= $8.8\times 0.48 J$

change in kinetic Energy = $\frac{1}{2}mv^2-0$

$8.8\times 0.48=\Delta K.E.$

K.E.-0= $4.22$

K.E.= $4.22\ J$

8 0

3 years ago

A(n) 930 N crate is being pushed across a level floor by a force of 400 N at an angle of 20◦ above the horizontal. The coefficie

Nana76 [90]

Answer:

The magnitude of the acceleration of the box is $2.291\,\frac{m}{s^{2}}$ .

Explanation:

The free body diagram of the crate is included as attachment, whose equations of equilibrium are described below:

$\Sigma F_{x} = P\cdot \cos 20^{\circ} - \mu_{k}\cdot N = \left(\frac{W}{g}\right)\cdot a$

$\Sigma F_{y} = P\cdot \sin 20^{\circ} + N - W = 0$

From second equation of equilibrium we find an expression for the normal force and find the respective value:

$N = W - P \cdot \sin 20^{\circ}$

$N = 930\,N - 400\cdot \sin 20^{\circ}\,N$

$N = 793.192\,N$

Lastly, the acceleration experimented by the crate during pushing is cleared in the first equation of equilibrium and consequently calculated:

$a = \frac{P\cdot \cos 20^{\circ}-\mu_{k}\cdot N}{\frac{W}{g} }$

$a = \frac{400\cdot \cos 20^{\circ}\,N-0.20\cdot (793.192\,N)}{\frac{930\,N}{9.807\,\frac{m}{s^{2}} } }$

$a = 2.291\,\frac{m}{s^{2}}$

The magnitude of the acceleration of the box is $2.291\,\frac{m}{s^{2}}$ .

3 0

3 years ago

What is the value of work done on an object when a 0.1x102–newton force moves it 30 meters and the angle between the force and t

Ganezh [65]

The Answer to the question above is A. 2.7 x 102 joules.

4 0

3 years ago

Read 2 more answers

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.8 km/h due east. Runner B is initia

vampirchik [111]

Answer:

0.39 km west

Explanation:

The position of Runner A is:

x = -5.7 + 8.8 t

The position of Runner B is:

x = 4.2 − 7.6 t

When the positions are equal:

-5.7 + 8.8 t = 4.2 − 7.6 t

16.4 t = 9.9

t = 0.604

Plug into either equation to find the position at this time:

x = -5.7 + 8.8 (0.604)

x = -0.39

The runners are 0.39 km west of the flagpole when they meet.

4 0

3 years ago

3. why is the sum of the maximum voltages across each element in a series r l c circuit usually greater than the maximum applied

Llana [10]

The sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

<h3>What is the Kichoff's loop rule?</h3>

Kirchhoff's loop rule states that the algebraic sum of potential differences, as well as the voltage supplied by the voltage sources and resistances, in any loop must be equal to zero.

In a series RLCcircuit, the voltages are not added by scalar addition but by vector addition.

Kirchhoff's loop rule is not violated since the voltages across different elements in the circuit are not at their maximum values.

Therefore, the sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

Learn more about Kichoff's loop rule at: https://brainly.in/question/35360816

#SPJ1

8 0

2 years ago