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Komok [63]
3 years ago
11

A 500-kilogram sports car accelerates uni-

Physics
1 answer:
timama [110]3 years ago
3 0

Answer:

90 meters

Explanation:

Given:

x₀ = 0 m

v₀ = 0 m/s

v = 30 m/s

t = 6 s

Find:

x

x = x₀ + ½ (v + v₀)t

x = 0 + ½ (30 + 0)(6)

x = 90

The car travels 90 meters.

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What does velocity tell about movement
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an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done
Andrei [34K]

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

Learn more about Work done here:

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7 0
1 year ago
What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?
ivann1987 [24]

Answer:

Explanation:The simple pendulum calculator finds the period and frequency of a ... Acceleration of gravity (g) ... Pendulum length (L) ... First of all, a simple pendulum is defined to be a point mass or bob (taking ... For example, it can be equal to 2 m. ... Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz

4 0
3 years ago
A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor a
Snezhnost [94]

Answer:

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t  eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5  eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6  eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

0.866(10.6) = 4.9t² + 1.1

9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

Finally, initial speed can be found by substituting the value of t into eq. 3

Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

8 0
3 years ago
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