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mariarad [96]
2 years ago
7

The body weighing 8kg moves in a straight line uniformly accelerated with an acceleration of 3m / s? on a horizontal surface, un

der the action of force 60N.Find a) What is the friction force acting on the body) The coefficient of friction between the body and the surface where it moves
Physics
1 answer:
Usimov [2.4K]2 years ago
3 0

Answer:

pls explain this to me

Explanation:

I will not answer it

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the coefficient of kinetic friction between a couch and the floor is 0.65. if the couch has a mass of 42 kg and you push it with
Anna [14]

Explanation:

The net force along the horizontal direction is

\sum{F_x} = F_{applied} - f

where f is the frictional force. We can find the frictional force by looking at the vertical forces acting on the couch:

\sum{F_y} = N - mg = 0 \Rightarrow N = mg

From the definition of frictional force,

f = \mu{N} = \mu{mg} = (0.65)(42\:\text{kg})(9.8\:\text{m/s}^2)

\:\:\:\:= 267.5\:\text{N}

Therefore, the net force on the couch is

\sum{F_x} = 600\:\text{N} - 267.5\:\text{kN} = 332.5\:\text{N}

4 0
2 years ago
If the velocity of a body varied uniformly from 10 m s-1 to 25 m s-1
BaLLatris [955]

Answer:

3m/s2

Explanation:

a = (v-u) / t

a = (25 - 10) / 5

a = 15 / 5

a = 3

4 0
2 years ago
What is the awnser to that question
Olegator [25]
The answers To your question is c
8 0
3 years ago
Read 2 more answers
An electric circuit can have no current when a switch is
Finger [1]

Answer:

Open

Explanation:

A switch is a part of a circuit where a connection can be made or broken. By convention, when the switch is "open", the connection is broken and current cannot pass. When the switch is "closed", the connection is complete and current can pass.

6 0
3 years ago
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
3 years ago
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