A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force

Answers

Answer 1

Answer:

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

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A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net work done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^

Answers

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= √((6.00m/s)^2+(2.00 m/s)^2)=6.32 m/s$
And so, the kinetic energy of the object is
$K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= √((8.00 m/s)^2+(4.00 m/s)^2)=8.94 m/s$
And so the new kinetic energy is
$K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

Final answer:

The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.

Explanation:

The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.

The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.

Learn more about Kinetic Energy and Work here:

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A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y-direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line. The electric field should be in the ______ direction.

Answers

Answer:

Negative z-direction.

Explanation:

We need to determine the direction of the magnetic force. Since the velocity of the proton is in the positive x direction, and the magnetic field is in the positive y direction, we know by the vectorial formula $F=q(v* B)$ (or, alternatively, with the left hand rule) that the magnetic force points in the positive z-direction (also taking into account that the charge is positive), so the electric field should be in the negative z-direction to balance it.

What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)

Answers

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=kma (where k is the proportionality constant)

k is 1 so you can ignore it.

So, our equation becomes F=ma

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Answers

Answer:

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Explanation:

Fb = Fg

qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6

= 0.115C

note:2800ft/sec = 853.44m/s

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , measured relative to the merry-go-round. neglect friction and the size of each child.

Answers

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

$m*2*R + Iw = 0$

$w = -(2mR)/(I)$

so merry go round will turn in opposite direction with above speed

A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g glass ball charged to 4.7 nC is shot straight up at 4.8 m/s from the floor level. How high does the ball go if the ceiling voltage is +3.0x10^6V?

Answers

The height at which the ball goes for the given parameters is; 0.827 m

What is the height of the ball?

We are given;

distance between the metal plates; d = 3.1 m

mass of glass; m = 1.1g = 0.0011 kg

charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C

speed of the glass ball; v = 4.8 m/s

voltage of the ceiling; V = +3.0 × 10⁶ V

The repulsive force experienced by the ball is gotten from the formula;

F = qV/d

|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1

|F| = 4.548 × 10⁻³ N

F = -4.548 × 10⁻³ N (negative because it is repulsive force)

The net horizontal force experienced by the ball is;

F_net = F - mg

F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)

F_net = -15.328 × 10⁻³ N

To get the height of the ball, we will use the formula;

F_net * h = ¹/₂mv²

h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)

We took the absolute value of F_net, hence it is not negative

h = 0.827 m

Read more about height of ball at; brainly.com/question/12446886

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

$F_(net) = F_c - mg\n\nF_(net) = -4.548 *10^(-3) - (1.1*10^(-3) * 9.8)\n\nF_(net) = -15.328*10^(-3) \ N$

The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.

$W = F_(net) *h\n\nW = 15.328 *10^(-3) * h$

W = K.E

$15.328*10^(-3) *h = (1)/(2)mv^2\n\n 15.328*10^(-3) *h = (1)/(2)(1.1*10^(-3))(4.8)^2\n\n 15.328*10^(-3) *h =0.0127\n\nh = (0.0127)/(15.328*10^(-3))\n\n h = 0.827 \ m$

Therefore, the ball traveled 0.827 m

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