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2 years ago

17 points

Physics

1 answer:

const2013 [10]2 years ago

3 0

The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>

MV = mv................. Equation 1

<h3>Where:</h3>

M = mass of the cannon
m = mass of the cannonball
V = velocity of the cannon
v = velocity of the cannonball

Make v the subject of the equation.

v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>

M = 500 kg
V = 3 m/s
m = 10 kg.

Substitute these values into equation 2.

v = (500×3)/10
v = 150 m/s.

Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

$L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K$

The view factor between two coaxial parallel disks would be

$\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33$

$\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75$

Then the view factor between base to top surface of the cylinder becomes $F_{12} = 0.26$ . From the summation rule

$F_{13} = 1-0.26$

$F_{13} = 0.74$

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

$\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}$

$\dot{Q_3} = 2\dot{Q_{13}}$

$\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)$

$\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)$

$\dot{Q_3} = 780.76W$

Therefore the rate heat radiation is 780.76W

5 0

3 years ago

Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,

madreJ [45]

Hello!

<em>a) What is the change in momentum</em>?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

$\boxed{a=\frac{V-Vi}{t} }$

Like you see, the final velocity will be zero, so:

$a = \dfrac{0m/s-40m/s}{4s}$

$a = \dfrac{-40m/s}{4s}$

$a = -10\ m/s^{2}$

Like you see, the aceleration applicate by superman is of <u>-10 m/s^2.</u>

If the aceleration is negattive, means the velocity will decrease.

<em>b.) What is the magnitud of the force wich superman exerts on the train?</em>

For calculate the force applicate for superman, lets applicate the second law of Newton:

$\boxed{F=ma}$

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of <u>-80000 Newtons.</u>

When the force is negative means the force applicated in the another direction of the object what is traveling.

8 0

3 years ago

What two quantities must stay the same in order for an object to have a constant velocity? A) The speed and kinetic energy must

GarryVolchara [31]

the answer is c) the speed and direction of travel must be constant

7 0

4 years ago

A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil

krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

Emf induced in the coil represented by formula

∈ = $-N\frac{d\phi}{dt}$ ...................(1)