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SpyIntel [72]
3 years ago
5

if a girl is running along a straight road with uniform velocity 1.5 metre per second find the acceleration​

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
4 0

Answer:

Acceleration is 0.

Explanation:

Since the girl is running with a uniform velocity, meaning she is moving at a constant rate, she is not accelerating. Acceleration only occurs when the speed of an object increases, decreases, or if there is a change in direction.

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outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
3 years ago
A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.
Vaselesa [24]
The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
                                     d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
                                     180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
                                      F = (1300 kg)(1.369 m/s²) 
                                            F = 1779.7 N
4 0
3 years ago
A man standing on top of a 30 m tall building throws a brick downwards with a velocity of 12 m/s. Determine the speed of the bri
loris [4]

Answer:

27.1m/s

Explanation:

Given parameters:

Height of the building  = 30m

Initial velocity  = 12m/s

Unknown:

Final velocity  = ?

Solution:

We apply one of the kinematics equation to solve this problem:

         v²  = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

          v²   = 12²  + (2 x 9.8 x 30)

          v  = 27.1m/s

3 0
3 years ago
A ball is thrown at a 60.0° angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the groun
klio [65]

Answer:

Option E is correct.

Time the ball remains in the air before striking the ground is closest to 3.64 s

Explanation:

yբ = yᵢ + uᵧt + gt²/2

yբ = 0

yᵢ = 2 m

uᵧ = u sinθ = 20 sin 60 = 17.32 m/s

g = -9.8 m/s², t = ?

0 = 2 + 17.32t - 4.9t²

4.9t² - 17.32t - 2 = 0

Solving the quadratic equation,

t = 3.647 s or t = -0.1112 s

time is a positive variable, hence, t = 3.647 s. Option E.

7 0
3 years ago
An object with acceleration of -10 m/s2 is ___.
aksik [14]

A. speeding up 10 m/s every second.


an acceleration with a negative number simply implies direction.

positive being north (for example) and negative being south

4 0
3 years ago
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