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3 years ago

Which sound characteristic is not affected by the relative motion of an object?

Physics

2 answers:

Tatiana [17]3 years ago

6 0

Answer:

intensity is correct

Explanation:

irinina [24]3 years ago

4 0

intensity is the answer. i just got it on the test

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Where does the word 'robot' come from?

Masja [62]

It comes the word robota in czech meaning drudgery or hard labor

8 0

3 years ago

The motion of an object is determined by the sum of the acting on it.

oee [108]

the force acting on that object

3 0

3 years ago

constant force F=5i+5j−1kF=5i+5j−1k is applied to an object that is moving along a straight line from the point (−5,−3,−4)(−5,−3

meriva

Answer:

W = 71J

Explanation:

Given force F = (5i+5j−1k)N

d = Δr

r1 = (−5,−3,−4)m

r2 = (2,5,0)m

Δr = r2 – r1 = (2-(-5), 5-(-3), 0-(-4))

Δr = (2+5, 5+3, 0+4) = (7i+ 8j +4k)m

W = F•d = (5i+5j−1k)•(7i+ 8j +4k)

W = 5×7 + 5×8 +-1×4 = 35 + 40 - 4

W = 71J

6 0

2 years ago

A long, thin solenoid has 700 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of

Aleks04 [339]

To solve this problem it is necessary to apply the concepts related to the magnetic field in a solenoid.

By definition we know that the magnetic field is,

$B = \mu_0 n I$

$\frac{dB}{dt} = \frac{d}{dt} \mu_0 NI$

$\frac{dB}{dt} = \mu_0 n\frac{dI}{dt}$

At the same tome we know that the induced voltage is defined as

$\epsilon = \frac{\Phi}{dt}$

$\epsilon = A \frac{dB}{dt}$

Replacing

$\epsilon = A \mu_0 n\frac{dI}{dt}$

$\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}$

PART A) Substituting with our values we have that

$\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0)(36)}{2}\\\epsilon = 0V/m\\$

Therefore there is not induced electric field at the center of solenoid.

PART B) Replacing the radius for 0.5cm

$\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0.5*10^{-2})(36)}{2}\\\epsilon = 7.9168*10^{-5}V/m$

Therefore the magnitude of the induced electric field at a point 0.5cm is $7.9168*10^{-5}V/m$

4 0

3 years ago

Read 2 more answers

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see

VMariaS [17]

Answer:

K = 4.64 10⁻¹⁵ J

Explanation:

For this exercise we can use conservation of energy

starting point. Next to the negative plate

Em₀ = U = e V

final point. Right when you hit the positive plate

Emf = K = ½ m v²

energy is conserved

Em₀ = Em_f

e V = K

The electric potential is related to the electric field

V = - Ed

we substitute

-e E d = K

let's calculate

K = -1.6 10⁻¹⁹ (-2.9 10⁶) 1.0 10⁻²

K = 4.64 10⁻¹⁵ J

7 0

2 years ago