Answer:
Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for 
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by on q. Then we can know the magnitude of the force exerted by 
about q, finally this will allow us to know the magnitude of
exerts a force on q in +y direction, and exerts a force on q in -y direction.
The net force on q is:
Rewriting for :
Answer:
The pressure exerted by camel feet is <u>2000 N/m²</u>.
Step-by-step explanation:
<h3><u>Solution</u> :</h3>
Here, we have given that ;
- Force applied on camel feet = 4000 N
- Total area of camel feet = 2 m²
We need to find the pressure exerted by camel feet.
As we know that :
Substituting all the given values in the formula to find the pressure exerted by camel feet.
Hence, the pressure exerted by camel feet is 2000 N/m².
Answer:
Explanation:
To solve this, we start by using one of the equations of motion. The very first one, in fact
1
V = U + at.
V = 0 + 0.8 * 3.4 = 2.72 m/s.
2.
V = 0 + 0.8 * 4.3 = 3.44 m/s.
3.
d = ½ * 0.8 * 4.3² + 3.44 * 12.9
d = 7.396 + 44.376
d = 51.77 m.
4.
d = 62 - 51.77 = 10.23 m. = Distance
traveled during deceleration.
a = (V² - Vo²) / 2d.
a = (0² - 3.44²) / 20.46
a = -11.8336 / 20.46 = -0.58 m/s²
5.
t = (V - Vo)/a =(0 - 3.44) / -0.58
t = -3.44/-.58 = 5.93 s
= Stop time.
T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total
time the hare was moving.
6.
d = Vo * t + ½ * a * t² = 62 m.
0 + 0.5 * (23.13)² * a = 61
267.5a = 61
a = 61/267.5
a = 0.23 m/s²
D. chemical to electrical to sound and electromagnetic
The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
To learn more about work done click on the given link brainly.com/question/25573309
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