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Rufina [12.5K]
3 years ago
7

Plzz answer this question correctly

Physics
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

by reducing friction.....

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You place the spring vertically with one end on the floor. You then lay a 1.60 kg book on top of the spring and release the book
malfutka [58]

Answer:

Compression in the spring, x = 3.7 cm

Explanation:

It is given that,

Mass of the book, m = 1.6 kg

It can be assumed the spring constant of the spring is, k = 840 N/m

As the book moves down, the change in potential energy of the book is converted to spring potential energy of compression. The mathematical expression is as follows :

\dfrac{1}{2}kx^2=mgx

x=\dfrac{2mg}{k}

x=\dfrac{2\times 1.6\ kg\times 9.8\ m/s^2}{840\ N/m}  

x = 0.037 meters

or

x = 3.7 cm

So, the spring is compressed to a distance of 3.7 cm. Hence, this is the required solution.

4 0
3 years ago
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0
3 years ago
A circular coil of wire of radius 5.0 cm has 20 turns and carries a current of 2.0 A. The coil lies in a magnetic field of magni
Korvikt [17]

Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A

That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m

7 0
3 years ago
he electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
Paraphin [41]

Answer:

R₁ = 50.77 Ω

Explanation:

Since, we know that:

Electric Power = P = VI

but from Ohm's Law:

V = IR

(or) I = V/R

Therefore,

P = V²/R

(OR) R = V²/P

where,

V = Battery Voltage

R = Resistance of combination

FOR SERIES COMBINATION:

R = Rs = (57 V)²/48 W

Rs = 67.69 Ω

but, we know that:

Rs = R₁ + R₂

R₁ + R₂ = 67.69 Ω

R₁ = 67.69 Ω - R₂  __________ eqn (1)

FOR PARALLEL COMBINATION:

R = Rp = (57 V)²/256 W

Rp = 12.69 Ω

but, we know that:

Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω

using eqn (1) and value of R₁ + R₂, we get

Rp = 12.69  = R₂(67.69 - R₂)/67.69

859.08 = 67.69 R₂ - R₂²

R₂² - 67.69 R₂ + 859.08 = 0

Solving this quadratic equation we get the answers:

Either, R₂ = 50.76 Ω

Either, R₂ = 16.92 Ω

Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,

<u>R₂ = 16.92 Ω</u>

using this value in eqn (1), we get:

R₁ = 67.69 Ω - 16.92 Ω

<u>R₁ = 50.77 Ω</u>

4 0
3 years ago
Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh
coldgirl [10]

Answer:

You cant draw on brainly

Explanation:

6 0
2 years ago
Read 2 more answers
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