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koban [17]
2 years ago
8

A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra

vel in the given time?
Physics
2 answers:
Natali5045456 [20]2 years ago
5 0
-- The truck begins the scenario moving at 6 m/s.

-- After increasing its speed at the rate of 2 m/s² for 12 seconds,
it has gained (2 x 12) = 24 m/s of speed, and it's then moving
at 30 m/s. 

-- The truck's average speed during the 12 seconds is (1/2) (6 + 30) = 18 m/s 

-- Traveling at an average speed of 18 m/s for 12 seconds,
the truck travels

(18 m/s) x (12 sec) = 216 meters  
SpyIntel [72]2 years ago
4 0

Answer:

216 meters i got the correct answer                

Explanation:

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Hi,

The correct answer is C. five 

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3 years ago
A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

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2 years ago
5. Radioactive isotopes can be used to find the age of rocks, fossils, or other artifacts. Carbon 14 has a half-life of 5,730 ye
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Answer:

A

Explanation:

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7 0
3 years ago
A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

4 0
2 years ago
The acceleration of an object depends upon
aalyn [17]
Its mass and net force acting on it
4 0
2 years ago
Read 2 more answers
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