Answer:
15 deg
Explanation:
Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦ .
Given:
For first ball, θ1 = 75◦
initial velocity for both the balls, u = 27.2 m/s
for second ball, θ2 = ?
since distance covered by both the balls is same.
Therefore,..
R1=(u^{2} sin2\alpha _{1}) /g[/tex]
the range for the first ball
the range for the second ball
R2=(u^{2} sin2\alpha _{2}) /g[/tex]
(u^{2} sin2\alpha _{2}) /g[/tex]=(u^{2} sin2\alpha _{1}) /g[/tex]
sin2\alpha _{2})=sin2\alpha _{1})
=sin^-1(sin2\alpha _{1})
=1/2sin^-1(sin2\alpha _{1})
=
15 deg
Answer:
C. Yes, the water could be changing the phase.
Explanation:
The best thing to do in this case is to redo the experiment and re record the info, it has to be precise and accurate so you also have to check if your procedure is correct. If the results are both accurate and precise then you have to report your findings to the committee of that specific field. <span />
Answer:
The second car must go with a speed of 63.43 m/sec
Explanation:
Speed V of lead car = 62.3 m/sec
Distance S = 55 laps = 55 ×400 meters=22000 m
We know
S = V × t
So,
t= S/V
We put values of S and V here, we get
t=22000/62.3
t= 353.1 sec
So in 353.1 sec the second car which is one lap behind - must go a distance of 55+1=56 laps or 56×400 m = 22400 meters to catch the lead car before it finishes.
i-e for second car
Distance S= 22400m
Time t = 353.1 sec
V= ?
using again
S=Vt
we get
V= S/t
V= 22400/353.1= 63.43 m/sec
