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2 years ago

14

what will be the final temperature is 0.05kg water at 0 degree centigrade is added to 0.25kg pf water at 90 degree centigrade sp

ecific heat capacity of water is 4200J/kgk

1 answer:

scoray [572]2 years ago

7 0

Q in = Q out

0.25 x 4200 x (90-t)=0.05 x 4200 x (t-0)

1050 (90-t)=210t

94500-1050t=210t

94500=1260t

t=75°

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A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/

Sauron [17]

Answer:

Part a)

$t = 3.85 s$

Part b)

$h = 72.67 m$

Part C)

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = -22.77 m/s$

Explanation:

Part a)

Horizontal speed of the cannon

$v = 30.0 m/s$

angle of projection

$\theta = 30^o$

now we have

horizontal speed = $v_x = vcos30 = 30 cos30 =25.98 m/s$

vertical speed = $v_y = vsin30 = 30 sin30 = 15 m/s$

now the time taken by it to cover the distance 100 m from the wall

$x = v_x t$

$100 = 25.98 t$

$t = 3.85 s$

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}(9.81)(3.85^2)$

$h = 72.67 m$

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

so we have

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = v_i + at$

$v_y = 15 - 9.81(3.85)$

$v_y = -22.77 m/s$

Part e)

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3 years ago

Read 2 more answers

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

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Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

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$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

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