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fenix001 [56]
3 years ago
14

a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

magnitude of the net force acting on the car during this 5.0 second interval
Physics
1 answer:
irinina [24]3 years ago
7 0
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


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A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?
belka [17]

Answer:

117.72kW

Explanation:

Given data

Mass m= 50kg

height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

WD= force * distance

WD= mgx

WD= 50*9.81*2

WD= 981 Joules

Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

Hence the power spent is 117.72kW

8 0
3 years ago
The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes
tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake  = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

4 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?
katrin [286]
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7 0
3 years ago
A 2.0kg rock is thrown straight up into the air with a speed of 30m/s. Ignore air resistance. What is the net force acting on th
bazaltina [42]

Answer:

19.6N

Explanation:

Given parameters:

Mass of rock = 2kg

Speed  = 30m/s

Unknown:

Net force on the rock  = ?

Solution:

The net force acting on this rock is a function of the acceleration due to gravity acting upon it.

 Net force  = weight  = mass x acceleration due to gravity

 Net force  = 2 x 9.8  = 19.6N downward

6 0
2 years ago
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