All categories

3 years ago

A ball is thrown off the top of a building and lands on the ground below.

Physics

1 answer:

natita [175]3 years ago

4 0

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

$K=\frac{1}{2} mv^2$

where $m$ is the mass of the object, and $v$ is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

mass of the ball
velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball before touching the gound can be determined.

You might be interested in

A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if

diamong [38]

Answer:

I believe that the answer is d.

Explanation:

Because there is nothing to make the aircraft accelerate or decelerate, it is going to stay in constant motion with no acceleration.

3 0

3 years ago

A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the

lilavasa [31]

Answer:

$R=803k\Omega$

Explanation:

We have the following information,

$V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s$

We apply the equation for capacitor charging the voltage across it,

$V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}$

Replacing values,

$R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}$

$R=803k\Omega$

3 0

3 years ago

At what speed must the electron revolve round the nucleus of

EleoNora [17]

Explanation:

I think this is it, give it a try

6 0

3 years ago

The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of

Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is $=4.51m/s$

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

Now to obtain the initial energy stored

Let U denote the initial energy stored and

$U = \frac{1}{2} kx^2$

Where x is the length the spring is displaced

k is the force constant of the string

$U = \frac{1}{2} * 627 * (0.6)^2$

$= 112.86 J$

Now referring to the formula above

i.e Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

$\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

$v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

$v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity $= 9.8m/s^2$ this displacement length of spring = 0.6

Therefore $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

$=4.51m/s$

8 0

3 years ago

The distance from the dwarf planet Pluto to the sun is 4.43 × 109 km at perihelion and 7.37 × 109 km at aphelion. Find the eccen

kaheart [24]

Answer:

0.249

Explanation:

Perihelion = 4.43 x 10^9 km

Aphelion = 7.37 x 10^9 km

Let e be the eccentricity and a be the length of semi major axis.

The relation between the semi major axis, perihelion and aphelion s given by

Semi major axis = half of sum of perihelion and aphelion

$a = \frac{4.43+7.37}{2}\times10^{9}$

a = 5.9 x 10^9 km

The relation between the perihelion, semi major axis and the eccentricity is given by

Perihelion = a (1 - e)

4.43 x 10^9 = 5.9 x 10^9 (1 - e)

0.751 = 1 - e

e = 0.249

3 0

3 years ago