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Anton [14]
3 years ago
7

Which of the following statements about the conservation of momentum is not correct? Momentum is conserved for a system of objec

ts pushing away from each other. The total momentum of a system of interacting objects remains constant regardless of forces between the objects. Momentum is conserved when two or more interacting objects push away from each other. Momentum is not conserved for a system of objects in a head-on collision.
Physics
1 answer:
Olin [163]3 years ago
3 0

Answer:

wrong statement :  Momentum is not conserved for a system of objects in a head-on collision.

Explanation:

In a head on collision of two objects , two equal and opposite forces are created at the point of collision . These two forces create two impulses in opposite direction which results in equal and opposite changes in momentum in each of them . Hence net change in momentum is zero. In this way momentum is conserved in head on collision of two objects.

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A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
Consider two waves X and Y traveling in the same medium. The two carry the same amount of energy per unit time, but X has one-se
RideAnS [48]

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy  = 1 / 7

λy / λx = 7 / 1

7 0
3 years ago
You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t
andrew-mc [135]

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

\beta_{dB} = 10log_{10} \frac{I}{I_0}

Where,

I = Acoustic intensity in linear scale

I_0 = Hearing threshold

The value in decibels is 17dB, then

17dB = 10log_{10} \frac{I}{I_0}

Using properties of logarithms we have,

\frac{17}{10} = log_{10} \frac{I}{I_0}

log_{10} \frac{I}{I_0} = 1.7

\frac{I}{I_0} = 10^{1.7}

\frac{I}{I_0} = 50.12 W/m^2

Therefore the factor that the intensity of the sound was 50.12W/m^2

5 0
3 years ago
Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative: (a)
Aleks04 [339]

Answer:

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

Explanation:

Work is defined by the expression

        W = F. r

bold letters indicate vectors, we can write this expression as a module

        W= F r cos θ

where is at the angle between force and displacement.

Let's apply this expression to the different cases

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

7 0
3 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
3 years ago
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