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snow_lady [41]
3 years ago
15

PLEASE HELP An automobile with an initial speed of 5.80 m/s accelerates uniformly at the rate of 4.1 m/s^2 Find the final speed

of the car after 5.6 s. Answer in units of m/s. Find the displacement of the car after 5.6 s. Answer in units of m.
Physics
2 answers:
Lemur [1.5K]3 years ago
8 0

Answer:

dfuvadfivu

Explanation:

e-lub [12.9K]3 years ago
6 0

Thanks for asking the question

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A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot
Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

3 0
2 years ago
In a convex lense f=20cm, m=1,then what is u and v?​
katen-ka-za [31]

Answer:

I'm not sure if I know whatever the answer is

3 0
2 years ago
A rocky space object of varying size
kow [346]
That would be an asteroid
3 0
3 years ago
A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces up
Serjik [45]

Answer:

0.906

Explanation:

Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second

\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW

So the efficiency of the water turbine is the ratio of output power over input power:

\frac{234}{258.307} = 0.906

3 0
3 years ago
7. How much work is done in moving a charge of 10 micro coulombs 1 meter along an equipotential of 10 volts?
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It takes work to push charge through a change of potential. 
There's no change of potential along an equipotential path,
so that path doesn't require any work.

8 0
3 years ago
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