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Yanka [14]
3 years ago
12

All the electrons within a magnetic substance are oriented

Physics
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:c

Explanation:

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Question 1 (1 point)
MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work W done by the frictional force is

W= Fd.

Now, F = 60N and d =10m; therefore,

W= (60N)(10m)

\boxed{W = 600J.}

Hence, the work done by friction is 660J.

7 0
3 years ago
The coefficient of kinetic friction between a suitcase and the floor is 2.76. if the suitcase has a mass of 71.5 kg, how far can
mihalych1998 [28]

suitcase can be pushed to a distance of 0.35 m

Explanation:

Work done= Ff d

w= 670 J

Ff= force of friction

d= distance

force of friction is given by Ff=μ mg

μ= coefficient of friction=2.76

m=mass=71.5 kg

so Ff= (2.76) (71.5) (9.8)

Ff=1934 N

so W= Ff d

670= 1934 d

d=0.35 m

8 0
3 years ago
Cuantos CM son 8 newtons
inn [45]

Answer: Be more specific

4 0
3 years ago
Read 2 more answers
Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​
Alinara [238K]

The volume of the room is the product of its dimensions:

10\times 12 \times 4 = 480\text{ m}^3

Now, from the equation

d=\dfrac{m}{V}

where d is the density, m is the mass and V is the volume, we deduce

m=dV

So, multiply the density and the volume to get the mass of air in the room.

7 0
4 years ago
How long will it take a 2.3"x10^3 kg truck to go from 22.2 m/s to a complete stop if acted on by a force of -1.26x10^4 N.What wo
Greeley [361]

The stopping distance is 45.0 m

Explanation:

First of all, we find the acceleration of the truck, by using Newton's second law:

F=ma

where

F=-1.26\cdot 10^4 N is the net force on the truck

m=2.3\cdot 10^3 kg is the mass of the truck

a is its acceleration

Solving for a,

a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2

where the negative sign means the acceleration is opposite to the direction of motion.

Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:

v^2-u^2=2as

where

v = 0 is the final velocity of the truck

u = 22.2 m/s is the initial velocity

a=-5.48 m/s^2 is the acceleration

s is the stopping distance

And solving for s,

s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
4 years ago
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