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3 years ago

What is the free-throw line used for?

Physics

1 answer:

Genrish500 [490]3 years ago

5 0

Answer:

it be the second one lad

Explanation:

i have done the playing pf the basketball

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Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n

damaskus [11]

Answer:

38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
Distance covered by car A = speed x time(t) = 60 x t = 60t
Distance covered by car B = speed x time(t) = 75 x t = 75t
60t + 75t = 340 miles
135t = 340
t = 2.51 hours
Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0

3 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

3 0

3 years ago

Why are dark matter and dark energy described as dark

zaharov [31]

Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it does not absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect.

3 0

2 years ago

A 2500 N force accelerates a car at a rate of 3.0 m/s^2. What is the car’s mass? 250 kg

Ronch [10]

Apply Newton's second law to the car's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

F = 2500N, a = 3.0m/s²

Plug in and solve for m:

2500 = m(3.0)

m = 830kg

5 0

3 years ago

A uniform steel of density 7800 kg/m³ weight 16g and is 250cm long .It is lengthens by 1.2mm when a force of 80N is applied.

Harlamova29_29 [7]

Explanation:

hopes it's helped uh mate!

4 0

2 years ago