Answer:42.43m/s
Explanation:According to vf=vi+at, we can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s
Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
Answer:
496.7 K
Explanation:
The efficiency of a Carnot engine is given by the equation:
where:
is the temperature of the hot reservoir
is the temperature of the cold reservoir
For the engine in the problem, we know that
is the efficiency
is the temperature of the cold reservoir
Solving for , we find:
Answer:
b.only when the current in the first coil changes.
Explanation:
An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.
Answer:
I think it is the Federal Pell Grant Program.
Explanation:
