A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

Ede4ka [16]

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

3 0

2 years ago

Bone has a Young’s modulus of about

MArishka [77]

Answer:

Explanation:

Bone has a Young’s modulus of about

1.8 × 1010 Pa . Under compression, it can

withstand a stress of about 1.66 × 108 Pa before breaking.

Assume that a femur (thigh bone) is 0.56 m

long, and calculate the amount of compression

this bone can withstand before breakin :).

6 0

3 years ago

Which dog has the most kinetic energy?

Nastasia [14]

Answer:

I'm pretty sure it'sssss A

4 0

2 years ago

A particle moving at speed 0. 32 c has momentum p0. the speed of the particle is increased to 0. 72 c. what is its momentum now?

Semenov [28]

A particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c then its momentum would be 2.25P₀.

<h3>What is momentum?</h3>

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

Mathematically the formula of the momentum is

P = mv

where P is the momentum of the particle

m is the mass of the particle

v is the velocity by which the particle is moving

As given in the question a particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c

by using the formula of the momentum and substituting the values of the velocity

P =mv

As mentioned in the question when the particle is moving with 0.32c velocity it has a momentum of P₀

P₀ = m*(0.32c)

If the speed of the particle is increased to 0.72 c the momentum would be

P=m*(0.72c)

by dividing the second equation from the first one

P₀/P = 0.32c/0.72c

P₀/P = 0.44

P =2.25P₀

Thus, the increased momentum of the particle would be 2.25P₀

Learn more about momentum from here

brainly.com/question/17662202

#SPJ4

6 0

1 year ago

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

TiliK225 [7]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .

As a result,

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V$ .

6 0

3 years ago