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NNADVOKAT [17]
3 years ago
6

What is the average power consumption in watts of an appliance that uses 5.00 KW.h of energy per day?

Physics
1 answer:
abruzzese [7]3 years ago
3 0
(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules

Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts

(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules
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Answer:
The Gravitational Force between the 2 masses is approximately 1.209x10^32 Newton’s

Explanation:

6 0
3 years ago
A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret
lukranit [14]

Explanation:

Given that,

Length of the spring, l = 50 cm = 0.5 m

Mass connected to the end, m = 330 g = 0.33 kg

The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm

(a) When the mass is connected, the force of gravity is balanced by the force in spring.

kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.15}\\\\k=21.56\ N/m

(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.

(c) The frequency of the oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{21.56}{0.33}} \\\\f=1.28\ Hz

Hence, this is the required solution.

7 0
4 years ago
What direct current in A will produce the same amount of thermal energy, in a particular resistor, as an alternating current tha
vodomira [7]

Answer:

I_{rms}= 4.314 AExplanation:According to the definition of the root mean square value of current, the value of direct current which passes through the same circuit in which an alternating current is passing and producing same amount of heat as the direct current produced is called root mean square value of current. So, the relation between the root mean square value of current and the peak value of alternating current is given by [tex]I_{rms}=\frac{I_{0}}{\sqrt{2}}

I_{rms}=\frac{6.1}}{\sqrt{2}}

[tex]I_{rms}= 4.314 A

6 0
3 years ago
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sec
faust18 [17]

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?

Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

7 0
3 years ago
A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before
Airida [17]

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so  W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times , hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J


5 0
4 years ago
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