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lapo4ka [179]
3 years ago
13

A truck traveling at a constant speed of 28 m/s passes a more slowly moving car. The instant the truck passes the car, the car b

egins to accelerate at a constant rate of 1.2 m/s^2 and passes the truck 545 m farther down the road. What was the speed of the car (in meters/second) when it was first passed by the truck?
Physics
1 answer:
choli [55]3 years ago
5 0

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck  = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

              time =\dfrac{distance}{speed}

              time =\dfrac{545}{28}

              time =19.46 s

velocity of the car when it crosses the truck

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

545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

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zimovet [89]

The gravitational force between the two balls is 6.13\cdot 10^{-9} N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem,  we have

m_1 = 6.8 kg

m_2 = 7.4 kg

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N

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4 0
3 years ago
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

Divide 1 & 2

\frac{5}{R_{moon}}=\frac{1}{6}

R_{moon}=30 m

4 0
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Concentrated solar power facilities are solar power—generating facilities that generate electricity at large centralized facilities and transmit that power to homes and businesses through the electric grid .

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Solar power refer to electric power or electricity that is generated from sun rays or radiations while using solar panels and other technologies.

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8 0
2 years ago
An object is placed to the left of a convex mirror, such that the object-to-image distance is 140 cm.
labwork [276]

Answer:

The focal length of the mirror is 52.5 cm.

Explanation:

Given that,

Object to Image distance d = 140 cm

Image distance v= 35 cm

We need to calculate the object distance

u = d-v

u = 140-35=105\ cm

We need to calculate the focal length

Using formula of mirror

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

Put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-105}+\dfrac{1}{35}

\dfrac{1}{f}=\dfrac{2}{105}

f=52.5\ cm

Hence, The focal length of the mirror is 52.5 cm.

6 0
3 years ago
What is the speed of a skater who travels a distance of 210 m in a time of 10 seconds?
Klio2033 [76]

21m per second. Take 210 divided by 10. Hope this helps!

7 0
3 years ago
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