Which of the diagrams is the correct model for the element beryllium? A) B) C) D)

A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang

MissTica

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

6 0

3 years ago

F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the

stepladder [879]

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

5 0

3 years ago

Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

τ = I α

T r = I α

T gR = I α

Alf = T R / I (1)

T = α I / R

Now let's use Newton's second law in the mass that descends

W- T = m a

a = (m g -T) / m

The two accelerations need related

a = R α

α = a / R

a = (m g - α I / R) / m

R α = g - α I /m R

α (R + I / mR) = g

α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

4 0

4 years ago

I WILL MARK BRAINLIEST!!ASAP!!! Wet Lab - Coulomb's Law lab from edge!!

anyanavicka [17]

Answer:

i don't get what I have to do

8 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago