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CaHeK987 [17]
3 years ago
5

Does anyone know the answer to this page and the next one that it has?

Physics
1 answer:
Gekata [30.6K]3 years ago
4 0
No sorry, wish I could help
You might be interested in
Let the resistance of an electrical component remain constant while the potential difference across the two ends of the componen
slamgirl [31]

Answer:

The current through it will also decrease to half of its former value because according to Ohm's law the current flowing through a resistor is directly proportional to the potential difference applied across its ends provided that the temperature and some other necessary conditions remain constant.

This is mathematically represented as follows;

V=IR.........(1)

The current is thus given as

I=\frac{V}{R}..............(2)

if R is constant and V is reduced to half, then we have the following;

I=\frac{V/2}{R}

Simplifying further we obtain

I=\frac{V}{2R}...........(3)

Equation (3) shows that the current I is also reduced to half.

8 0
3 years ago
Read 2 more answers
A 3.90 kg block is in equilibrium on an incline of 31.0◦. The acceleration of gravity is 9.81 m/s2 . What is Fn of the incline o
storchak [24]

Answer:

Explanation:

The sum of the pore along the plane is expressed according to Newton's law

Fn-Ff = ma

Fn is the moving force

Ff = nR = frictional force

m is the Mass

a is the acceleration

Substitute the given values

Fn - nR = ma

Fn - tan31°(mgcostheta) =3.9(9.8)

Fn - tan31(3.9(9.8)cos31) = 3.9(9.8)

Fn - tan31(38.22cos31)= 38.22

Fn - 32.76tan31 = 38.22

Fn-19.68 = 38.22

Fn = 38.22+19.68.

Fn = 57.90N

Hence Fn (moving force) of the inclined block is 57.90

4 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0
3 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s
AlexFokin [52]

Answer:

K=0.023J

Explanation:

From the question we are told that:

Mass m=0.15

Velocity v=0.5m/s

Angular Velocity \omega=8.4rad/s

Generally the equation for Kinetic Energy is mathematically given by

K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)

K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))

K=0.023J

8 0
2 years ago
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