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lukranit [14]
2 years ago
11

A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co

me to rest in 4.0 m?
Physics
1 answer:
Rus_ich [418]2 years ago
3 0

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N

So, the required force is 319.2 N.

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A truck traveling down the highway collides with a slower moving mosquito traveling in the same direction. Which of the followin
Ipatiy [6.2K]

Answer:

B. The truck and mosquito exert the same size force on each other.

Explanation:

Newton's third law (law of action-reaction) states that

"When an object A exerts a force (action) on an object B, then object B exerts an equal and opposite force (reaction) on object A"

In this case, we can call

object A = the truck

object B = the mosquito

Thereforce according to Newton's third law, the force exerted by the truck on the mosquito is equal in magnitude to the force exerted by the mosquito on the truck (and in opposite direction).

The reason for which the mosquito will experience much more damage is the fact that the mosquito's mass is much smaller than the truck's mass, and since the acceleration is inversely proportional to the mass:

a=\frac{F}{m}

the mosquito will experience a much larger deceleration than the truck, therefore much more damage.

6 0
2 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
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Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

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3 years ago
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What is the resultant force on the human cannonball in the
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Answer:

heart

Explanation:

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2 years ago
Lionel makes a graphic organizer to compare the electric field around a positive charge with the electric field around a negativ
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