<h2><em>Answer: According to wikipedia Sir Isaac Newton has told this statement.</em></h2><h2><em>hope its helps you.</em></h2><h2><em>have a great day.</em></h2><h2><em> keep smiling be happy stay safe. </em></h2><h2><em /></h2>

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7 0

2 years ago

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
$F=12N$

Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
$m = \frac{36.0}{6.0}$
$\therefore \: m = 6.0kg$

You can also use the equation of linear motion,
$v = u + at$
$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
$F=ma$
$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
$\therefore \: m = 6kg$

4 0

2 years ago

A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

Lisa [10]

Answer:

Vertical distance= 3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time= 0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

3 0

2 years ago