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2 years ago

6

A horizontal force of 50 N causes a trolley to move a horizontal distance of 30M How much work is done on the trolley by the for

ce?

1 answer:

zepelin [54]2 years ago

3 0

Answer:

1500J

Explanation:

w=f×s

50×30

=1500j

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Jessica is running a 10K. She alternates between running and walking each kilometer. She runs at a rate of 1 kilometer every 5 m

saveliy_v [14]

Answer:

c.100 minutes

Explanation:

Total distance = 10 km

Runs for 1 km every 5 minutes

walks 1 km every 15 min

She alternates between walking and running so, Jessica will walk 5 km and run 5 Km

Time taken by Jessica for walking : 5 km

Time taken to walk 1 km=5 minutes

Time taken to walk 5 km

=> 5 X 5

=>25 minutes

Time taken by Jessica for Running : 5km

Time taken to run 1 km = 15 minutes

=> 5 X 15

=>75 minutes

Total time taken = Time taken by Jessica for walking + Time taken by Jessica for Running

=>25 minutes +75 minutes

=> 100 minutes

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2 years ago

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What is the gravitational field theory and how does it work? four mark answer. Don't give me random answers just for points or i

MrMuchimi

It is a theory on a show that people try to solve.

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2 years ago

Which car has the most kinetic energy ??

gregori [183]

Kinetic Energy is defined by 1/2 times the mass times the velocity squared.

KE = (1/2)mv2

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2 years ago

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if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a

Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

4 0

2 years ago

A particle with a charge of -4.0 μC and a mass of 3.2 x 10-6 kg is released from rest at point A and accelerates toward point B,

Assoli18 [71]

Answer:

$V_B-V_A=-20736-0=-20736volt$

Explanation:

We have given charge on the particle $q=-4\mu C=-4\times 10^{-6}C$

Mass of the charge particle $m=3.2\times 10^{-6}kg$

From energy of conservation kinetic energy will be equal to potential energy

So at point A

$\frac{1}{2}mv^2=qV$

At point a velocity is zero

So $\frac{1}{2}(3.2\times10^{-6} )0^2=-4\times 10^{-6}V_a$

$V_A=0volt$

At point B velocity will be 72 m/sec

So $\frac{1}{2}\times 3.2\times 10^{-6}72^2=-4\times 10^{-6}V_b$

$V_B=-20736volt$

So $V_B-V_A=-20736-0=-20736volt$

4 0

2 years ago