What is the difference in the speed of sound on a warm day versus on a cold day?

Physics

1 answer:

seraphim [82]3 years ago

4 0

The sun is bright and when its cold there is no sun

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A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

6 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$