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choli [55]
3 years ago
13

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

s the car goes through a dip with a 40.0 radius of curvature.
Physics
1 answer:
Rashid [163]3 years ago
8 0

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

A = W + \dfrac{W}{2}

A =\dfrac{3W}{2}

A =\dfrac{3mg}{2}

When the car is at the bottom,  the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

N = m g + \dfrac{mv^2}{r}

\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}

\dfrac{mg}{2} = \dfrac{mv^2}{r}

v = \sqrt{\dfrac{rg}{2}}

v = \sqrt{\dfrac{40\times 9.8}{2}}

v = 14 m/s

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They both have a certain force. They are different because that force is different
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Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat
aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

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rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R    ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

5 0
3 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0
3 years ago
a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time
Evgen [1.6K]

Answer:

1.43 s

Explanation:

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The vertical distance covered by an object in free fall is given by

S=ut + \frac{1}{2}at^2

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

S=\frac{1}{2}gt^2

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s

3 0
3 years ago
Which law of motion accounts for the following statement?
julia-pushkina [17]
That would be the second law
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