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3 years ago

Which car has the most kinetic energy? (PLEASE HELP NEED ANSWER ASAP)

Physics

1 answer:

Hatshy [7]3 years ago

6 0

Kinetic energy = 1/2 mv²

so, you will have to do it for every single one of them if you want steps.

Kinetic Energy = For A, 202500J

For B, 90000J

For C, 168750J

For D, 75000J

From the above we an say that kinetic energy is highest in option A.

In case, there are no steps needed, just by the values we are able to understand the answer as in Option A, eventhough the mass is the same, higher the velocity or speed, higher the kinetic energy

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When two equal forces act on the same object in opposite directions, what is the net force?

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Answer:

Explanation:

Forces with equal magnitudes and opposite directions cancel each other out, so the net force is 0.

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3 years ago

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When the mass of an object decreases, the force of gravity - Remains Unchanged - Decreases - increases - Becomes irregular

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The force of gravity decreases

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Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

3 years ago

On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1

Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m = W/g

Where:

W : Weight (N)

g : acceleration due to gravity (m/s²)

Data

W₁ =1.8 kN : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁ m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}$

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}$

a₂= -2.17 m/s² : acceleration of the worker

5 0

3 years ago

Ali is whirling a 2.0 kg bunch of bananas in a circular path having a radius of 0.50 m. The bananas complete 2 revolutions every

skad [1K]

Answer:

1) 2.467 N

2) a) 0.248m

b) 2.3π rad/sec

Explanation:

Given data:

mass of Banana bunch ( m ) = 2.0 kg

radius of circular path ( R ) = 0.5 m

number of revolutions completed = 2

Time to complete 2 revolutions = 6 seconds

1) Determine the force to keep the motion constant for one complete revolution in every 4 seconds

F = mv^2 / r ----- ( 1 )

where V = 2πR/T

where : R = 0.5 , m = 2, T = 4 seconds

Insert values into equation 1

F = 2 * 4π^2 * 0.5/4^2

= 2.467 N

2a) Calculate the maximum distance of coin from center

angular velocity ( w ) = v/r

coefficient of static friction ( μ ) = 0.25

$F_{c} = u mg$ ---- ( 1 )

mv^2/r = μmg --- ( 2 ) cancelling the mass on both sides eqn 2 becomes

v^2 = μ*g*r

dividing both sides of equation by r^2

w^2 = μ*g/r

hence determine distance ( r ) of coin from center

r = 0.25 * 9.81 / π^2 = 0.248 m

2b ) determine the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping

distance coin is placed ( r ) = 4.7 cm = 0.047 m

find speed of rotation ( w )

w^2 = μ*g/r

w = √ 0.25 * 9.81/ 0.047

= 7.2236 rad/secs ≈ 2.3π rad/sec

3 0

2 years ago