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3 years ago

What is the equivalent of 0° C in Kelvin? Help will give brainiest

Physics

1 answer:

givi [52]3 years ago

8 0

Answer:

273.15 dat is the answer im pretty sure

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the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago

The velocity of an object includes its speed and what

SCORPION-xisa [38]

I believe that the correct answer you are looking for is the distance traveled

5 0

3 years ago

Read 2 more answers

Which contribution did Kepler make to viewing the solar system?

amm1812

He proposed the sun-centered model of the solar system.

8 0

3 years ago

Which of these statements about the sun's energy is true? (5 points)

s344n2d4d5 [400]

It is transferred evenly to water through air

7 0

3 years ago

(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new

Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation

$m \frac{d^{2}x}{dt^{2}} +kx=0$

Definition of parameters

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

$x(t)=c1cos(ωt)+c2Sin(ωt)$

$x(t)=c1cos(5.929t)+c2Sin(5.929t)$

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is

x(t)=0.337sin((5.929t)

6 0

3 years ago