PHYSICS COLLEGE

A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?

Answers

Answer 1

Answer:

Answer:

a) x = 5.48 10⁻² m and b) 0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

A = √ (2Em / k)

A = √ (2 0.12 / 80)

A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

Em = K = ½ m v²

Final

Em = Ke = ½ k x²

½ m v² = ½ k x²

x = √(m/k) v

x = 2 √(0.50 /800.0)

x = 0.05 m

Answer 2

Answer:

Answer:

a) The greatest extension of the spring is 0.055 m

b) The spring compress 0.05 m

Explanation:

Please look at the solution in the attached Word file

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How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 J
B) 5 J
C) 50 J
D) 1 J
E) 10 J

Answers

Answer:

option C

Explanation:

given,

Force on the object = 10 N

distance of push = 5 m

Work done = ?

we know,

work done is equal to Force into displacement.

W = F . s

W = 10 x 5

W = 50 J

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

Final answer:

The work done on an object when a force of 10 N pushes it 5 m is 50 Joules, calculated by multiplying the force and the displacement. So, the correct option is C.

Explanation:

The question is asking about work, which in physics is the result of a force causing a displacement. The formula for work is defined as the product of the force (in Newtons) and the displacement (in meters) the force causes. If a force of 10 N pushes an object a distance of 5 m, the work done is calculated by multiplying the force and the displacement (10 N * 5 m), yielding 50 Joules of work.

Therefore, the correct answer is 50 J (C).

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What unit is used to measure energy when calculating specific heat capacity? Give the abbreviation, not the full name.

Answers

The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

What unit is used to measure energy?

The specific heat capacity should be determined in joules per kilogram degree-celsius ( J k g − 1 ∘ C − 1 ).

It is to be supplied for the substance with respect to mass and it increased the temperature.

Hence, The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

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The unit used to measure energy when calculating specificheat capacity is the joule (J).

The joule (J), a unit used to quantify energy, is used to calculate specific heat capacity. In the International System of Units (SI), the joule serves as the default unit of energy.

It is described as the quantity of energy that is delivered when one newton of force is exerted across a one-meter distance.

Thus, the quantity of heat energy needed to increase the temperature of a particular substance by a specific amount is measured as specificheat capacity. J/kg°C, or joules per kilogramme per degree Celsius, is the unit of measurement.

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Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?

Answers

Explanation:

The relation between resistance and resistivity is given by :

$R=\rho (l)/(A)$

$\rho$ is resistivity of material

l is length of wire

A is area of cross section of wire

Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is

A^^\-> points in the -x direction with a magnitude of 21. What is the y component of A^^\->

Answers

Answer:

$A_y=-36.37^(\circ)$

Explanation:

Given that,

Vector A points in the -x direction with a magnitude of 21.

Let the x component is making an angle of 60 degrees with negative x axis. The x component of a vector is given by :

$A_x=A\ cos\theta$

$-21=A\ cos(60)$

$A=(-21)/(cos(60))$

A = -42 units

The y component of a vector is given by :

$A_y=A\ sin\theta$

$A_y=-42\ sin(60)$

$A_y=-36.37^(\circ)$

So, the y component of vector A is (-36.37) degrees. Hence, this is the required solution.

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?

Answers

A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $(2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) ) = (106x)/(x+53)$

$26.5 = (106x)/(x+53) \n \n 79.5 x = 1404.5\n \n x = 17.67 miles/hour$

) A circular coil of diameter 20. cm, with 16. turns is in a 0.13 Tesla field. (a) Find the total flux through the coil when the field is perpendicular to the coil plane. (b) If the coil is rotated in 10. ms so its plane is parallel to the field, find the average induced emf.

Answers

Answer:

(a) 0.0041 weber

(b) 0.41 volt

Explanation:

diameter of coil, d = 20 cm

radius of coil, r = half of diameter = 10 cm = 0.1 m

magnetic field strength, B = 0.13 tesla

(a)

The angle between the normal of the coil and the magnetic field is 0°.

Magnetic flux, Ф = B x A x Cos 0°

Ф = 0.13 x 3.14 x 0.1 x 0.1 x 1

Ф = 0.0041 Weber

(b)

angle between the magnetic field and the normal of the coil is 90°.

time, t = 10 ms = 0.01 s

final flux = B x A x cos 90° = 0

induced emf = rate of change of magnetic flux

e = (0.0041 - 0) / 0.01

e = 0.41 Volt

Answer:

a) $\phi=0.4084\ T.m^2$

b) $emf=653.44\ V$

Explanation:

Given:

diameter of the coil, $d=20\ cm=0.2\ m$

no. of turns in the coil, $N=16$

magnetic field strength to which the coil is subjected, $B=0.13\ T$

time taken by the coil to rotate from normal the field to parallel, $t=10* 10^(-3)\ s$

The flux through the coil can be given as:

$\phi=BA$

where:

$A=$ area enclosed by the section of the coil

$\phi=0.13* \pi* (0.2^2)/(4)$

$\phi=0.4084\ T.m^2$

When the coil is rotated there is change in flux which lead to an induced emf in the coil according to the Faraday's law:

$emf=N(d\phi)/(t)$

where:

$d\phi=$ change in the flux

here the flux changes from maximum value to zero when the coil becomes parallel to the field lines because then there is no field line intercepting the coil area.

$emf=16* (0.4084)/(0.01)$

$emf=653.44\ V$

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