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salantis [7]
3 years ago
6

I WILL GIVE BRAINLIEST TO WHO ANSWERS CORRECTLY

Physics
2 answers:
Jlenok [28]3 years ago
5 0

Dynamic equilibrium.

REY [17]3 years ago
3 0

Answer:

Static Equilibrium

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N discussing engines, the ratio of output work to input work expressed as a percentage is called
djyliett [7]

The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.

5 0
3 years ago
A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 4.50 T dir
notka56 [123]

Answer:

39.375 A

Explanation:

To find the induced current, we use the relation

e = -ΔΦ/Δt, where

ΔΦ = change in magnetic flux of the bracelet

Δt = change in time, = 20 ms

Also, Φ = A.ΔB, such that

A = area of the bracelet, 0.005m²

ΔB = magnetic field strength of the bracelet = 1.35 - 4.5 = -3.15 T

ΔΦ = A.ΔB

ΔΦ = 0.005 * -3.15

ΔΦ = -.01575 wb

e = -ΔΦ/Δt

e = -0.01575 / 20*10^-3

e = 0.7875 V

From the question, the resistance of the bracelet is 0.02 ohm, so

From Ohms Law, I = V/R

I = 0.7875 / 0.02

I = 39.375 A

6 0
3 years ago
The potential energy of an apple is 6.0 joules. The apple is 3m high. What is the mass of the apple?
zalisa [80]

Answer:

0.204 kg

Explanation:

6/(9.8*3)

6/29.4=0.204

9.8 represents gravity

5 0
3 years ago
A 300 g bird flying along at 6.0 m/s sees a 10 g insect heading straight toward it with a speed of 30 m/s. the bird opens its mo
Bumek [7]

Answer:

<em>6.77m/s</em>

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the velocities before collision

v is the final collision

Given

m1 = 300g  = 0.3kg

u1 = 6.0m/s

m2 = 10g = 0.01kg

u2 = 30m/s

Required

The bird's speed immediately after swallowing v

Substitute the given values into the formula

m1u1 + m2u2 = (m1+m2)v

0.3(6) + 0.01(30) = (0.3+0.01)v

1.8+0.3 = 0.31v

2.1 = 0.31v

v = 2.1/0.31

<em>v = 6.77m/s</em>

<em>Hence the bird's speed immediately after swallowing is 6.77m/s</em>

5 0
3 years ago
A gas sample is heated from -20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L. If the initial pressure is 0.14
Svetach [21]

Answer:

The answer is e.

Explanation:

We take:

T_{1}=253K

V_{1}=2.00 l

P_{1}=0.14atm

V_{2}=4.50 l

T_{2}=330K

Taking the gas as an ideal gas, we can use the ideal gas law:

\frac{PV}{nRT}

⇒ n=\frac{P_{1}V_{1}}{RT_{1}} ⇒ n=0.013mol

Then:

P_{2}=\frac{nRT_{2}}{V_{2}} ⇒ P_{2}=0.0811 atm

Taking R=0.08205atmL/molK.

5 0
3 years ago
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