<h2>
<em><u>⇒</u></em>Answer:</h2>
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Step-by-Step Solution:
Solution 35PE
This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.
Step 1 of 3<
/p>
The legs have an extension of 0.600 m in the crouch position.
So, m
The person is at rest initially, so the initial velocity will be zero.
The acceleration is m/s2
Acceleration m/s2
Let the final velocity be .
Step 2 of 3<
/p>
Substitute the above given values in the kinematic equation ,
m/s
Therefore, the final velocity or jumping speed is m/s
Explanation:
Answer:
A 'kink' in the glass tube which breaks the mercury as it contracts, storing the highest temperature reading. The glass tube is shaped like a lens to magnify the thin mercury thread. Shaking the thermometer resets the mercury back into the bulb.
Answer:
u = 104.68 m/s
Explanation:
given,
horizontal distance = 150 m
elevation of 12.4 m
angle = 8.6°
horizontal motion = x = u cos θ. t .............(1)
vertical motion =
................(2)
from equation(1) and (2)
..........{3}
u = 104.68 m/s
The initial speed of the ball is u = 104.68 m/s
Answer: 24
Explanation:
Given the following equation:
Where 
is the number of mushrooms in a pizza and 
the number of pizzas.
If we know the restaurant will make 8 pizzas (
), then:
This is the needed number of mushrooms for 8 pizzas
Answer:
2.145×10^-10 V or 0.2145nV
Explanation:
From hf=eV
h= Plank's constant = 6.6×10^-34JS
f= frequency of the electromagnetic wave = 5.2×10^4 Hz
e= electronic charge= 1.6×10^-19 C
V= voltage
V= hf/e
V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C
V= 2.145×10^-10 V or 0.2145nV
Therefore the voltage created is 2.145×10^-10 V or 0.2145nV
