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umka21 [38]
3 years ago
7

when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe

n the rate is 52 cm?
Physics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

Rate of change of area will be 9.796cm^2/min

Explanation:

We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

Radius of the circular plate r = 52 cm

Area is given by A=\pi r^2

So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

So rate of change of area will be 9.796cm^2/min

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Name the organs of Movement in​
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An earthworm moves by using two different sets of muscles. 2. FISHES :- Fish swim by flexing their bodies and tail back and forth. Fish stretch or expand their muscles on one side of their body, while relaxing the muscles on the other side.

3 0
3 years ago
A heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h. I
Natali5045456 [20]

The power that heat pump draws when running will be 6.55 kj/kg

A heat pump is a device that uses the refrigeration cycle to transfer thermal energy from the outside to heat a building (or a portion of a structure).

Given a heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h and if the COP of the heat pump is 2.8

We have to determine the power the heat pump draws when running.

To solve this question we have to assume that the heat pump is at steady state

Let,

Q₁ = 22000 kj/kg

COP = 2.8

Since heat pump used to heat a house runs about one-third of the time.

So,

Q₁ = 3(22000) = 66000 kj/kg

We known the formula for cop of heat pump which is as follow:

COP = Q₁/ω

2.8 = 66000 / ω

ω = 66000 / 2.8

ω = 6.66 kj/kg

Hence the power that heat pump draws when running will be 6.55 kj/kg

Learn more about heat pump here :

brainly.com/question/1042914

#SPJ4

5 0
2 years ago
The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.0 cm apart with a 20 kV potential
Julli [10]

Answer:

A) electric field strength between the plates;E = 2 x 10^(6) N/C

B) exit velocity;v = 8.39 x 10^(7) m/s

Explanation:

We are given;

Potential difference; V = 20 kV = 20000 V

Distance between the 2 parallel plates; d = 1cm = 0.01 m

A) The electric field strength will be gotten from;

E = V/d

E = 20000/0.01

E = 2000000

E = 2 x 10^(6) N/C

B) For exit speed, we'll use the formula for Kinetic energy; KE = (1/2)mv²

KE is also expressed as; V•q_e

Thus,

(1/2)mv² = V•q_e

Where;

V is potential difference = 20000 V

Q_e is charge of electron which has a constant value of; (1.6 x 10^(-19))C

m is mass of electron with a constant value of (9.1 x 10^(-31)) kg

v is the velocity

Thus, making v the subject, we have;

v = √((2V•q_e)/m)

v = √((2 x 20000•(1.6 x 10^(-19)))/(9.1 x 10^(-31)))

v = 83862786 m/s or

v = 8.39 x 10^(7) m/s

3 0
3 years ago
An object is moving with constant non-zero velocity. Which of thw following statements about it must be true
hram777 [196]

Answer:

The net force on the object is zero.

Explanation:

An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".

5 0
3 years ago
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