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Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad

dimaraw [331]

When looking at it from a physics standpoint, the thick padding decreases the amount of force that a player experiences. Force equals mass times acceleration where acceleration is distance over time. The thick padding increases the amount of time the player makes contact with the post which decreases force.

8 0

3 years ago

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A substance is known to reflect red and blue light. What color would it have when it is illuminated by green light?

oksian1 [2.3K]

Answer:

A substance known to reflect red and blue light, and illuminated by green light, will have all the three primary light colors. The color it would reflect is white.

Explanation:

In visible light spectrum, a combination of red light and blue light gives magenta. Also a combination of red and green lights gives yellow light. and when green and blue lights mix the resulting light color is cyan.

However, if these three primary colors mix in equal proportion, we will have a neural color or white color and in the absence of these primary colors we will have black.

Thus, a substance known to reflect red and blue light, and illuminated by green light, will have all the three primary light colors. The color it would reflect is white.

7 0

3 years ago

Lars is standing near the edge of a 90-meter cliff. He throws a ball upward, but does not catch it, and it falls to the bottom o

Kipish [7]

Answer:

c. 43 m/s

Explanation:

Given the following data;

Displacement, S = 90 meters

Time, t = 5.55 seconds

To find the initial velocity;

We would use the second equation of motion given by the formula;

$S = ut + \frac {1}{2}at^{2}$

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

We know that acceleration due to gravity is -9.8m/s² because the direction is downward.

Substituting into the equation, we have;

$90 = u*5.55 + \frac {1}{2}*(-9.8)*5.55^{2}$

$90 = u5.55 - 4.9*30.8025$

$90 = u5.55 - 150.93225$

Rearranging the equation, we have;

$u5.55 = 90 + 150.93225$

$u5.55 = 240.93225$

$u = \frac {240.93225}{5.55}$

Initial velocity, u = 43.41 ≈ 41 m/s

4 0

3 years ago

If an alpha particle and a beta particle have the same energy, which particle will penetrate farther into an object?

lidiya [134]

Answer:

Beta particle

Explanation:

If an alpha particle and a beta particle have the same energy, beta particle will penetrate farther into an object than alpha particle because;

1. In air, beta particles can travel a few hundred times farther than alpha particles.

2. Beta particles have more penetrating power than alpha particles.

3. Alpha particle can be absorbed or stopped by an inch or less 1-2 centimeters of air or a thin piece of tissue while beta particles can be stopped or absorbed by a thin layer of Aluminium.

4. Alpha particles loose all of their energies in a small volume easily that beta particles.

5. Beta particles can traveled a longer distance between 2-3 meters at a speed nearly equal to that of light than alpha particles which travel a distance between 2-4 cm at a speed approximately five percent the speed of light.

5 0

3 years ago

I’M DESPERATE PLZ HELP!! 40 POINTS!!

Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, $F_k$ , required is given by the formula, F = m × a,

From which we have;

$F_k$ = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, $F_k$ = 0.0625 N

c. The coefficient of friction between the ice and puck, $\mu_k$ , is given from the equation for the kinetic friction force as follows;

$F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8$

$\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755$

The coefficient of friction between the ice and puck, $\mu_k$ ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force $F_k = 0.012755 \times 1 \times 9.8 = 0.125 \ N$

The acceleration, a = $F_k$ /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here, $F_k = 0.125 \ N$

The coefficient of friction $\mu_k$ ≈ 0.012755 is constant

5 0

3 years ago