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2 years ago

✫ state the work - energy theorem. thankyou ~

Physics

2 answers:

artcher [175]2 years ago

6 0

Answer:

The work-energy theorem states that the work done by the net force acting on a body is equal to the change produced in the kinetic energy of the body.

Please mark as Brainliest!!

KengaRu [80]2 years ago

5 0

$2as = 2 \frac{f}{m} s = {v}^{2} - {v0}^{2}$

$w = fs = \frac{1}{2} m {v}^{2} - \frac{1}{2} m {v0}^{2}$

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

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Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is

goblinko [34]

Answer:

d $\frac{x}{l}$ = m× λ⇒ d = λ ×m×l / x

= 630× $10^{-9}$ m × 3×3m/ 45× $10^{-3}$ m

= 1.26× $10^{-4}$ m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0

3 years ago

a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the

aleksklad [387]

At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

 H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

 H = 20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

 50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add (1/2 G Q²) to each side.
Then it says
 50 = 20Q

Divide each side by 20: 2.5 = Q .

And there we are. The stones pass each other

 2.5 seconds

after they are simultaneously launched.

5 0

3 years ago

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.

Then for this case the relationship becomes: $x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m$ .

Note that the acceleration is negative because is a braking process.

4 0

3 years ago

Read 2 more answers

What is precision?what is precision?

RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

6 0

2 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago

✫ state the work - energy theorem. thankyou ~​

✫ state the work - energy theorem. thankyou ~