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Hoochie [10]
3 years ago
6

The acronym laser stands for light amplification by ____ emission of radiation

Physics
1 answer:
AleksAgata [21]3 years ago
6 0
<span>light amplification by stimulated emission of radiation </span>
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A ball is kicked from the top of a building with a velocity of 50 m/s and lands 165 m away from the base of the building.
Nadusha1986 [10]

The height of the building is 53.4 m (projectile motion)

Explanation:

The motion of the ball is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

First of all, we consider the horizontal motion, which is a uniform motion at constant speed. The equation that gives the horizontal distance travelled is:

d=v_x t  

where:

v_x = 50 m/s is the horizontal velocity of the ball, which is constant since there are no forces in this direction

d = 165 m is the horizontal distance travelled by the ball

Solving for t, we find the time of flight:

t=\frac{d}{v_x}=\frac{165}{50}=3.3 s

Now we analyze the vertical motion: it is an accelerated motion, we can use the suvat equation

s=ut+\frac{1}{2}at^2  

where  

s is the vertical displacement, which is the height of the building

u = 0 is the initial vertical velocity of the ball

t = 3.3 s is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity  

Solving for s, we find the height of the building:

s=0+\frac{1}{2}(9.8)(3.3)^2=53.4 m

Learn more about projectile motion here:

brainly.com/question/8751410

#LearnwithBrainly

6 0
4 years ago
A seated cable row is an example of which level of training in the NASM OPT model?
Vera_Pavlovna [14]

Answer:

B. Strength

Explanation:

The OPT Model, or Optimum Performance Training Model, is a "fitness training system created by the NASM. The OPT Model is contructed with scientific evidence and principles that progresses an individual through five training phases: stabilization endurance, strength endurance, hypertrophy, maximal strength and power".

On the stabilization level we have the phase 1 called stabilization endurance.

For the level strength part we have 3 phases . Phase 2: Strength endurance , Phase 3: Hypertrophy, Phase 4: Maximal strength. And we can consider the case "A seated cable row" on this the level strength since we need to have some abilities to do this but not enough to stay on the power level since this one is the advancd level.

For the power level we have the last phase called power in order to mantain and conduct high training level programs.

8 0
4 years ago
A 4000-? resistor is connected across a 220-V power source. What current will flow through the resistor?
Varvara68 [4.7K]

Answer:

55 mA

Explanation:

Ohm's law states:

V = IR

where V is voltage, I is current, and R is resistance.

220 V = I (4000 Ω)

I = 0.055 A

I = 55 mA

7 0
3 years ago
Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc
Vanyuwa [196]

Answer:

1\cdot 10^{-8} N to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

q_1=2nC=2\cdot 10^{-9}C is the magnitude of charge 1

q_2=5nC =5\cdot 10^{-9}C is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0
3 years ago
A boy pushes on a wagon so that it accelerates at a rate of 0.50 m/s2. The wagon has a mass of 24 kg. What is the magnitude of t
IRISSAK [1]

Answer:

12N

Explanation:

Force acting on any body is equal to mass of body multiplied by acceleration produced in the body.

Its formula is

F = M*a

where F is the force acting on body

M is the mass of the body

a is the acceleration produced in the body

Unit of force is Newton represented by N

1 N is force acting on body of mass 1 kg which produces acceleration of 1m/s^2.

_____________________________________

Given

a = 0.50 m/s^2

M = 24 kg

therefore

F = 24*0.5 = 12 N

Thus,  magnitude of the boy's pushing force is 12 N.

4 0
4 years ago
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