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2 years ago

What is the free-throw line used for?

Physics

1 answer:

Genrish500 [490]2 years ago

5 0

Answer:

it be the second one lad

Explanation:

i have done the playing pf the basketball

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A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

Thus, the value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0

2 years ago

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

What are the units of acceleration?

Marina86 [1]

Acceleration is any change in speed or direction of motion.

The dimension of speed is [length/time],
so a change is [length/time²].

Popular units include [meter/second²] and [feet/second²] .
________________________

Direction almost always boils down to an angle, (which technically
has no dimensions), so a change in direction is [angle/time] .

Popular units include [radian/second] and [degree/second] .

5 0

3 years ago

Partially correct answer iconYour answer is partially correct. A proton initially has and then 2.30 s later has (in meters per s

morpeh [17]

Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.

a) = (1.82i - 9.69k)m/s2

b) magnitude = 9.85m/s2

c) direction = 280.64 degree

Explanation:

The detailed and step is shown in the attachment.

7 0

3 years ago

Determine the direction of the force that will act on the charge in each of the following situations. A negative charge moving t

wlad13 [49]

Answer:

a) DOWN direction, b) directed INTO THE SCREEN, c) F = 0

Explanation:

The direction of the force is

for electric force

F = q E

where we assume a positive test charge, for which the force has the direction of the electric field.

For a magnetic field

in this case the direction of the force is given by the right hand rule.

For a positive test charge, the thumb points in the direction of velocity, the other fingers extended in the direction of the magnetic field, and the palm gives the direction of force for a positive charge.

F = q v x B

Let us apply these considerations to our case.

a) negative charge moving to the left

in a magnetic field points away from the screen

In this case the thumb goes to the left, the fingers extended outwards and the palm points upwards, but since the charge is negative the force has a DOWN direction.

b) negative charge moves to the left

in electric field it points off the screen.

The outside is in the direction of the electric field and since the charge is negative, the force is directed INTO THE SCREEN

c) positive charge moves down

in magnetic field points up

in this case the velocity and the field have the same direction so the vector product of them is zero

F = q v B sin 0

F = 0

6 0

3 years ago