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gulaghasi [49]
2 years ago
11

Which of the following are for vector directione?

Physics
1 answer:
Inga [223]2 years ago
8 0
Answer:
C. outside 45 degrees
Step-by-step explanation:
C. outside 45 degrees
because it can be any point or direction outside 45 degrees
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The third law of thermodynamics,the principle of temperature.

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At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m
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1) The car overtakes the truck at a distance of 160 m far from the intersection.

2) The velocity of the car is 40 m/s

Explanation:

1)

The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by

x_c(t) = \frac{1}{2}at^2

where

a=5 m/s^2 is the acceleration

t is the time

On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by

x_t(t) = vt

where

v = 20 m/s is the velocity of the truck

t is the time

The car overtakes the truck when the two positions are the  same, so when

x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s

So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is

x_c(8) = \frac{1}{2}(5)(8)^2=160 m

So, the car overtakes the truck 160 m far from the intersection.

2)

The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

For the car in this problem, we have:

u = 0 (it starts  from rest)

a=5 m/s^2

And we know that the car overtakes the truck when

t = 8 s

Substituting into the equation,

v=0+(5)(8)=40 m/s

Learn more about accelerated motion:

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3 years ago
Phineas and Ferb are riding their bikes on Main Street. They begin at 11 Mile Rd at
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Explanation:

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A toy rocket fires its engines to launch it straight up from rest. After a short time, the engine turns off and the
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at 10 seconds.. b/c that's when the velocity stops increasing..

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To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
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