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MakcuM [25]
3 years ago
14

A bus travels north on some busy city streets for 2.5 km, and a trip

Physics
1 answer:
d1i1m1o1n [39]3 years ago
8 0

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

Hence, the velocity of the  bus, V = 11.31 m/s

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Samantha and Emily are pushing a box of textbooks in the same direction across their classroom. Samantha is applying a force of
velikii [3]

Answer: 20 newtons

Explanation:

Given that:

Force applied by Samantha = 10 newtons

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Net force of their efforts = ?

Since net force of forces applied in the same direction is obtained by adding up the seperate forces applied, then

Net force = (10 newtons + 10 newtons)

Net force = 20 newtons

Thus, the net force of their efforts is 20 newtons.

3 0
3 years ago
What is the speed of light with N=1.33
alexgriva [62]

If the refractive index of some substance is  1.33, then
the speed of light in that substance is
                  
               (speed of light in vacuum) / (1.33)  = 

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3 0
3 years ago
Read 2 more answers
In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls
sweet-ann [11.9K]

Answer:

Acceleration=0.5m/s^2

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

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Using the formula

Acceleration of box=\frac{20}{40}=0.5m/s^2

The acceleration of the box=0.5m/s^2

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

a=\frac{30}{40}=\frac{3}{4} ms^{-2}

v^2-u^2=2as

Substitute the values

v^2-0=2\times \frac{3}{4}\times 0.3=0.45

v^2=0.45

v=\sqrt{0.45}=0.67m/s

Hence, the speed of the box after it has  been pulled a distance of 0.3 m=0.67 m/s

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2 years ago
If the load on Pulley E is 20N, how much effort is needed to life it?
Mariana [72]
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3 years ago
Drag each tile to the correct location.
Lady_Fox [76]

The order of the location of each process in the carbon cycle are as follows;

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