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:
The work done by gravity is 784 J.
Explanation:
Given:
Mass of the block is, 
Height to which it is raised is, 
Acceleration due to gravity is, 
Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,
Force of gravity is given as the product of mass and acceleration due to gravity.
. Now,
Therefore, the work done by gravity is 784 J.
Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s
Here
- Acceleration and initial velocities are constant.
According to first equation of kinematics.
- Time was t at velocity v
- Time will be 4t at velocity 4v
Answer: 0°
Explanation:
Step 1: Squaring the given equation and simplifying it
Let θ be the angle between a and b.
Given: a+b=c
Squaring on both sides:
... (a+b) . (a+b) = c.c
> |a|² + |b|² + 2(a.b) = |c|²
> |a|² + |b|² + 2|a| |b| cos 0 = |c|²
a.b = |a| |b| cos 0]
We are also given;
|a+|b| = |c|
Squaring above equation
> |a|² + |b|² + 2|a| |b| = |c|²
Step 2: Comparing the equations:
Comparing eq( insert: small n)(1) and (2)
We get, cos 0 = 1
> 0 = 0°
Final answer: 0°
[Reminders: every letters in here has an arrow above on it]
