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kvv77 [185]
2 years ago
13

Which of the following is NOT a product of science?

Physics
2 answers:
emmasim [6.3K]2 years ago
5 0
Answer: D. None of these
Marina86 [1]2 years ago
3 0

Answer:

D, None of these.

Explanation:

They have science in them but its not a product.

You might be interested in
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
What distance does electromagnetic radiation travel in 55.0 μs ?
Komok [63]
Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
D = ( 3 x 10^8 m/s ) (  55 x 10^-6 s )
D = 16500 m
4 0
3 years ago
4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav
Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

                                                                         = 4.5 x 10⁹ Km

                                                                          = 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

                                           <em>V = d / t    m/s</em>

∴                                           t = d / V

                                               = 4.5 x 10¹¹ m / 3 x 10⁸ m/s

                                               = 15,000 s

                                               = 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

4 0
3 years ago
a 2,000 pound car is driving at 60 miles/hour along a straight, level road. what is the net force acting on the car?
SVETLANKA909090 [29]

Answer:

0

Explanation:

According to Newton's second law, the net force is equal to the mass times the acceleration.  Since the car is not accelerating, the net force is 0.

5 0
3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
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