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2 years ago

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The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

?

2 answers:

luda_lava [24]2 years ago

5 0

Answer:

56J

Explanation:

Formula to calculate kinetic energy:

$K=(1/2)*m*v^{2}$ ,equation (1)

K: It's the kinetic energy in Joules (J)

m: body mass in kilograms (kg)

v: body speed in meters / second (m / s)

Known information:

m = 7kg

v = 4 m / s

We replace the known information in the equation 1:

$K=( 1/2)*7*4^{2}$

$K=56 kg*m^{2} /s^{2} = 14 Newton*metro$

$K=56J$

Lisa [10]2 years ago

3 0

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

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What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p

Ray Of Light [21]

Answer:

A. Volume is unchanged

Explanation:

$P_{i}$ = initial pressure of the gas = 722 torr = 96258.7 pa

$P_{f}$ = final pressure of the gas = 0.950 atm = 96258.75 pa

$T_{i}$ = initial temperature = 32 °F = 272.15 K

$T_{f}$ = final temperature = 273 K

$V_{i}$ = initial volume

$V_{f}$ = final volume

Using the Equation

$\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}$

Inserting the values

$\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}$

$V_{f} = (1.00312) V_{i}$

Hence the volume is unchanged.

4 0

2 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)

Where:

. $\phi = BAcos\theta$ { B is magnetic field }

{A is cross-sectional area}

. $N =$ No. of turns in coil.

. $\frac{d\phi}{dt} =$ Rate change of induced Emf.

Here,

Considering the case :-

$N1 = 2N$ & $\frac{d\phi1}{dt} = 2\frac{d\phi}{dt}$

Putting these value in the equation (1) and finding the new emf induced (∈1)

∈1 = $-N1\times\frac{d\phi1}{dt}$

∈1 = $-2N\times2\frac{d\phi}{dt}$

∈1 = $4 [-N\times\frac{d\phi}{dt}]$

∈1 = 4∈ ...............{from Equation (1)}

Hence,

The Resultant Induced Emf in coil is 4∈.

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2 years ago

A physics teacher is designing a ballistics event for a science competition. The ceiling is 3.00m high, and the maximum velocity

krek1111 [17]

The launch proyectiles of kinematics allows to find the maximum initial vertical velocity of the body so that it just reaches the ceiling

v_{oy} = 2.56 m / s

Given parameters

The ceiling height y = 3 m

To find

Maximum vertical speed

Projectile launching is an application of kinematics where on the x axis there is no acceleration or and on the y axis the acceleration is the acceleration of gravity (g = 9.8 m / s ^ 2)

In this case, the maximum vertical velocity that the body can have occurs when the velocity on the ceiling is zero.

v_y² = v_{oy}² - 2 g y

where v and v_{oy} are the initial velocity at the ceiling e initial, respectively, g the acceleration of gravity e and the height

0 = v_{oy}² - 2 g y

v_{oy} = $\sqrt{2gy}$

v_{oy} = $\sqrt{2 \ 9.8 \ 3}$

v_{oy} = 2.56 m / s

In conclusion with the kinematic of launch projectiles we can find the maximum initial vertical velocity of the body so that it just reaches the ceiling

v_{oy} = 2.56 m / s

learn more about projectile launch here:

brainly.com/question/10903823

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1 year ago

Calculate the amount of energy produced in a nuclear reaction in which the mass defect is 0.187456 amu.

luda_lava [24]

For nuclear reactions, we determine the energy dissipated from the process from the Theory of relativity wherein energy is equal to the mass defect times the speed of light. We calculate as follows:

E = mc^2 = 0.187456 (3x10^8)^2 = 1.687x10^16 J

Hope this answers the question.

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2 years ago

Read 2 more answers

A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force F is exerted tangentially to th

Stella [2.4K]

Given :

Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.

A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².

To Find :

The magnitude of F.

Solution :

Torque on hoop is given by :

$\tau =F\times R\\\\I\alpha = FR\\\\MR^2\alpha = FR\\\\F = MR\alpha$ ( Moment of Inertia of hoop is MR² )

Putting value of M, R and α in above equation, we get :

$F=5\times 2\times 2.5\ N\\\\F = 25 \ N$

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.

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1 year ago