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2 years ago

What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months

Physics

1 answer:

iren2701 [21]2 years ago

7 0

Answer:

93312000 Joules

Explanation:

Energy can be explained as the capacity for doing work. And can be expressed as

E= pt....................eqn(1)

Where

E= Energy in Joules

P= power in Watt= 2.00W

t= time in seconds

Time= 18 months, which can be converted to "seconds" as;

18 months = [ 18 monthsx 30 days x 24 hours x 60 min x 60 sec] = 46656000

Then

time= 46656000 seconds

Power=2 watt

If we substitute into equation (1) we have

Energy = 2 watt × 46656000 seconds

Energy= 93312000 Joules

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Jalen is sliding around on his kitchen floor in his

tresset_1 [31]

Answer:

his feet feel warmer because of friction. when friction takes place then an object will warm up because friction is a universal law that always takes place. if you rub your hands together, you also experience the heat of frictional I hope that helped.

5 0

3 years ago

Motion equation of object is

Ganezh [65]

Take the derivative to find the velocity of the object:

$v=\dfrac{\mathrm dx}{\mathrm dt}=-12+6t$

The object stops when $v=0$ :

$-12+6t=0\implies6t=12\implies t=2$

so the answer is E.

5 0

3 years ago

Two students were climbing stairs at school. Student 1 has a weight of 700 N. Student 2 has a weight of 650 N. How much power wo

KIM [24]

Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

From the question,

We are to calculate the power each student would have to climb the flight of stairs.

Power can be calculated using the formula

$P = \frac{F \times d}{t}$

Where

P is Power

F is the force

d is the distance

and t is the time

NOTE: The weight of the students represent the force

For student 1

F = 700 N

d = 4 m

t = 6 s

∴ $P = \frac{700 \times 4}{6}$

P = 467 W

For student 2

F = 650 N

d = 4 m

t = 6 s

∴ $P = \frac{650 \times 4}{6}$

P = 433 W

Hence, Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

Learn more here: brainly.com/question/18801566

3 0

2 years ago

A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel

lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: $-10 [N] = m.ax$

Y-direction balance force: $-m*9,8 \frac{m}{s^2} = m.ay$

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

$ax=(-10 [N]) / m$

$ay=-m*9,8\frac{m}{s^2} / m$

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

$ay=-9,8\frac{m}{s^2}$

For the x component aceleration we must replace the Newton unit:

$N =\frac{kg.m}{s^2}$

$ax= -10 \frac{kg.m}{s^2} / (2 kg)$

$ax= - 5 \frac{m}{s^2}$

6 0

3 years ago

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$