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2 years ago

A ball is thrown vertically upward with an initial velocity

Physics

1 answer:

ivolga24 [154]2 years ago

5 0

Answer:

Explanation:

First we define our variables

V0=29.4

a=-9.8

V=0

We have to find the maximum displacement , which I will define as X

We use formula v^2=v0^2+2aX

All we do is substitute our values

0=29.4^2-19.6X

29.4^2=19.6X

X=29.4^2/19.6=44.1

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for the same type of atom with 8 protons, 9 neutrons, and 10 electrons, what type of atom or ion is it

Ilia_Sergeevich [38]

Answer:

O²⁻

Explanation:

Number of protons = 8

Number of neutrons = 9

Number of electrons = 10

What type of atom or ion is it = ?

Solution:

Protons are the positively charged particle in an atom

Neutrons do not carry any charges

Electrons are negatively charged particles

For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.

Now,

Charge = Number of protons - Number of electrons

Charge = 8 - 10 = -2

The charge on the atom is -2 and so it is an oxygen ion with -2 charge

The ion is O²⁻

8 0

2 years ago

A police radar gun uses X-band microwave radiation at a frequency of 12.2 GHz. Microwaves travel at the speed of light, or 3x108

quester [9]

Answer:

A police radar gun uses X-band microwave radiation at a frequency of 13.1 GHz. Microwaves travel at the speed of light, or 3x108 m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car.

a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

ANSWER: 2533 Hz

Explanation:

6 0

3 years ago

A 9,100-microfarad [muF] capacitor is charged to 22 volts [V]. If the capacitor is completely discharged through an iron rod

MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0

3 years ago

You want to move a 4- kg bookcase to a different place in the living room. If u push with a force of 65 n and the bookcase accel

IrinaK [193]

Answer:

1.65

Explanation:

The equation of the forces along the horizontal direction is:

$F-F_f = ma$ (1)

where

F = 65 N is the force applied with the push

$F_f$ is the frictional force

m = 4 kg is the mass

$a=0.12 m/s^2$ is the acceleration

The force of friction can be written as $F_f = \mu R$ (2), where

$\mu$ is the coefficient of kinetic friction

R is the normal force exerted by the floor

The equation of forces along the vertical direction is

$R-mg=0$ (3)

since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find

$F-\mu mg = ma$

And solving for $\mu$ ,

$\mu = \frac{F-ma}{mg}=\frac{65-(4)(0.12)}{4(9.8)}=1.65$

7 0

2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago