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bezimeni [28]
3 years ago
5

A girl and her bicycle have a total mads of 42 kg. At the top of the hill her speed is 4 m/s. The hill is 14.3m high and 112m lo

ng. If the force of friction as she rides down the hill is 20N, what us the speed at the bottom.
Physics
1 answer:
frosja888 [35]3 years ago
4 0

Answer:

13.78 m/s

Explanation:

Given that:

The mass of the girl & her bicycle = 42 kg

The speed at the top of the hilll= 4 m/s

The height of the hill = 14.3 m

The length of the hill (distance) = 112 m

The frictional force = 20 N

To find the speed at the bottom of the hill; we need to carry out the following processes.

The workdone by gravity = mass × acceleration due to  gravity × Δh

= 42 × 9.8 × ( 14.3 - 0 )

= 5885.88 joules

The workdone by the friction = Force × distance = - 20 × 112    (since she is riding down the hill)

The workdone by the friction = -2240 Joules

The initial Kinetic friction = 1/2 mv²

= 1/2 × 42 × 4²

= 336 Joules

The final kinetic energy = Initial Kinetic energy + total work  

The final kinetic energy = (336 + 5885.88 - 2240) Joules

The final kinetic energy = 3981.88   Joules

Using the final kinetic energy =   1/2 mv²

3981.88  = 1/2 × 42 × v²

3981.88  = 21 v²

v² = 3981.88/21

v² =189.61

v = \sqrt{189.61}

v = 13.78 m/s

Therefore, the speed at the bottom = 13.78 m/s

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7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm  (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b.  1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm  (Approx)

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5 0
3 years ago
I need to find the current resistance and voltage for each in this complicated circuit plz help
konstantin123 [22]

Explanation:

The 11Ω, 22Ω, and 33Ω resistors are in parallel.  That combination is in series with the 4Ω and 10Ω resistors.

The net resistance is:

R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)

R = 20Ω

Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:

V = IR

120 V = I (20Ω)

I = 6 A

So the voltage drops are:

V = (4Ω) (6A) = 24 V

V = (10Ω) (6A) = 60 V

That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:

V = 120 V − 24 V − 60 V

V = 36 V

So the currents are:

I = 36 V / 11 Ω = 3.27 A

I = 36 V / 22 Ω = 1.64 A

I = 36 V / 33 Ω = 1.09 A

If we wanted to, we could also show this using Kirchhoff's laws.

7 0
3 years ago
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