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VLD [36.1K]
3 years ago
14

Hi i just need some points and i want to give some to you

Physics
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

ok thanks

Explanation:

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A series RL circuit includes a 6.05 V 6.05 V battery, a resistance of R = 0.655 Ω , R=0.655 Ω, and an inductance of L = 2.55 H.
Ivenika [448]

Answer:

The induced emf 1.43 s after the circuit is closed is 4.19 V

Explanation:

The current equation in LR circuit is :

I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })    .....(1)

Here I is current, V is source voltage, R is resistance, L is inductance and t is time.

The induced emf is determine by the equation :

V_{e}=L\frac{dI}{dt}

Differentiating equation (1) with respect to time and put in above equation.

V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }

V_{e}=Ve^{\frac{-Rt}{L} }

Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.

V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }

V_{e}=4.19\ V

5 0
3 years ago
Compute the work (in joules) required to compress a spring 3 cm
faust18 [17]

Answer:

7×160=1120kj so,1120kj=1.12j.

Explanation:

Energy is equal to force times distance,so we add all the distance and we multiply them with the total force of the spring to get the answer. However, the answer is in kilojoules so we have to change it in to joules by dividing the answer into joules by 1000..

5 0
3 years ago
If the forces acting on an system produce a net force, what can we say about these forces
lutik1710 [3]
Net force refers to the (vector) sum of all the forces acting on something. It's a mathematical construction, so it's not a single identifiable force. The forces themselves are real, but the net force is not an actual force.

Hope this helps you out.
8 0
3 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
Precision
Schach [20]

Answer:

how close a true value a measurement is

6 0
3 years ago
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