Answer:
2.605m
Explanation:
Using the formula for calculating Range (distance travelled in horizontal direction)
Range R = U√2H/g
U is the speed = 4.8m/s
H is the maximum height = ?
g is the acc due to gravity = 9.8m/s²
R = 3.5m
Substitute into the formula and get H
3.5 = 4.8√2H/9.8
3.5/4.8 = √2H/9.8
0.7292 = √2H/9.8
square both sides
0.7292² = 2H/9.8
2H = 0.7292² * 9.8
2H = 5.21
H = 5.21/2
H = 2.605m
Hence the height of the ball from the ground is 2.605m
Answer:
88 m/s
Explanation:
To solve the problem, we can use the following SUVAT equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance covered
For the car in this problem, we have
d = 484 m is the stopping distance
v = 0 is the final velocity
is the acceleration
Solving for u, we find the initial velocity:
Answer:
t = 0.319 s
Explanation:
With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by
v = √T/λ
Linear density is
λ = m / L
λ = 4/20
λ = 0.2 kg / m
The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg
T = W = mg
T = 80 9.8
T = 784 N
The pulse rate is
v = √(784 / 0.2)
v = 62.6 m / s
The time it takes to reach the hook can be searched with kinematics
v = x / t
t = x / v
t = 20 / 62.6
t = 0.319 s
Answer:
c) Very dangerous and users must not override devices designed to protect from exposures.
Explanation:
X-ray is a form of high energy electromagnetic radiation and are part of the electromagnetic spectrum.
X-ray radiation from diffraction and fluorescence instruments is very dangerous because of their high energy and wavelength.
Hence, users must not override devices designed to protect from exposures. The best shielding device to protect one from exposure is Lead.
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