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sertanlavr [38]
3 years ago
14

The air in your classroom has

Physics
1 answer:
mel-nik [20]3 years ago
8 0
D.) all of the above
You might be interested in
When a coil is carrying a current of 25.0 A that is increasing at 145 A/s the induced emf in the coil has magnitude 3.70 mV.
Aleks04 [339]

Answer:

a) 25.5 µH

b) 22.95 mV

Explanation:

Induced emf in a inductor is given by

E = L * di/dt, where

E is the voltage of the circuit

L is the inductance of the circuit

di/dt if the rate of inductance

A

So we have

0.0037 = L * 145

L = 0.0037 / 145

L = 0.0000255

L = 25.5 µH

B

i(t) = 225t²

Recall that

E = L * di/dt, so that

E = 25.5 µH * |225t²|

Differentiating with respect to t, we have

E = 25.5 * 2 * 225t

E = 25.5 * 450t

Solving for t = 2,we get

E = 25.5 * 450(2)

E = 25.5 * 900

E = 22950 µV or

E = 22.95 mV

6 0
3 years ago
Solve 13.004+3.09+112.947
Mashcka [7]

Answer:

The answer is 129.041

Explanation:

Because when you add desimals you need to keep them lined up and not uneven

4 0
3 years ago
Tính hiệu suất nhiệt của động cơ nhiệt biết nhiệt lượng ở nguồn nóng 420,4kJ/kg và nhiệt lượng ở nguồn lạnh 218kJ/kg.
Travka [436]

Answer:

69

Explanation:

just because

6 0
3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m
forsale [732]

Answer:

Acceleration due to gravity will be g=5.718m/sec^2

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing =\frac{145}{100}=1.45sec

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

T=2\pi \sqrt{\frac{l}{g}}

So 1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}

\sqrt{\frac{0.55}{g}}=0.230

Squaring both side

{\frac{0.3025}{g}}=0.0529

g=5.718m/sec^2

So acceleration due to gravity will be g=5.718m/sec^2

3 0
3 years ago
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Anit [1.1K]

Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

Explanation:

Gravitational Force Fg = GMm/r2----1

Where G is gravitational constant

M Mass of the planet, m mass of the orbit and r is the distance between the masses.

Since the circular orbit move around the planet, it means they do not touch each other.

The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.

Total distance r= (R1 + d + R2)^2---2

Recall, density rho =

Mass M/Volume V

Hence, mass of planet = rho × V

But volume of a sphere is 4/3πr3

Therefore,

Mass M of planet = rho × 4/3πr3

=4/3πr3rho in kg

From equation 1 and 2

Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

6 0
3 years ago
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