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shutvik [7]
3 years ago
12

Calculate the amount of work for a car that exerts 10N and travels a distance of 5 meters.

Physics
2 answers:
lesya [120]3 years ago
4 0

Answer:

50 j

Explanation:

W=Force*distance

W=10N*5M

W=50 NM

n*m = joules so you can say 50joules

Andreas93 [3]3 years ago
4 0
50 J


Explanation......



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A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest
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Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

3 0
3 years ago
the engine of a car of amass of 2000 kg produced a force of 15000N find the acceleration of the car​
weqwewe [10]
I believe the answer would be 7.5 m/s^2
3 0
3 years ago
A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w
Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0
3 years ago
Read 2 more answers
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
Transmission Lines and Health. Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the elec
Blababa [14]

Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:

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Given values:

μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m

Plug in and solve for B:

B = 4π×10⁻⁷(170)/(2π(8.0))

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4.25×10⁻⁶/(5.0×10⁻⁵)

= 0.085

= 8.5%

The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.

8 0
4 years ago
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