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Alla [95]
3 years ago
7

Driving your car with a constant speed of 11 m/s, you encounter a hump in the road with a circular cross section. If the radius

of curvature of the hump is 54 m, what is the apparent weight of a 80 kg person in your car as you pass over the top of the hump?
Physics
1 answer:
Simora [160]3 years ago
6 0

Answer:604.74  N

Explanation:

Given

Speed of car=11 m/s

radius of curvature is 54 m

mass of person=80 kg

At Top

mg-N=\frac{mv^2}{r}

where N=normal reaction

N=mg-\frac{mv^2}{r}

N=80\time 9.8-\frac{80\times 11^2}{54}

N=604.74 N

Apparent weight will be equal to Normal reaction

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Brilliant_brown [7]
Hey there! 

The correct answer to your question is: Intensity

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7 0
2 years ago
Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat
nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

d = 48 km

now the time taken by the boat to move to and fro is given as

t = \frac{d}{v}

t = \frac{48 + 48}{48}

t = 2 hrs

Time taken by Boat B to cover the distance

t = \frac{48}{24} + \frac{48}{72}

t = 2.66 h

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

3 0
3 years ago
A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame
Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

f=\dfrac{v}{2l}

v is the speed of the mass

Speed is given by :

v=\sqrt{\dfrac{T}{\mu}}

\mu is the mass per unit length,\mu=\dfrac{m}{l}

f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}

T is the tension in the wire,

f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}

T=4f^2lm

T=4(120)^2\times 0.75\times 0.012

T = 518.4 N

Tension in the wire, T = m' g

m'=\dfrac{T}{g}

m'=\dfrac{518.4}{9.8}

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

6 0
3 years ago
Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f
Step2247 [10]
The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.
8 0
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When several radio telescopes are wired together, the resulting network is called a radio
WINSTONCH [101]

Answer:

Interferometer

Explanation:

4 0
3 years ago
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