Answer:
a
b
Explanation:
From the question we are told that
The length of the Chinook salmon is 
The mass of the Chinook salmon is 
The upward velocity in water is 
The upward velocity in air is 
Generally from kinematic equations
=> ![5.80^2 = 3.00^2 + 2 * a * [ [\frac{2}{3} * l ]](https://tex.z-dn.net/?f=%205.80%5E2%20%20%3D%20%203.00%5E2%20%2B%20%202%20%20%2A%20%20a%20%2A%20%20%5B%20%5B%5Cfrac%7B2%7D%7B3%7D%20%2A%20l%20%5D)
=> ![5.80^2 = 3.00^2 + 2 * a * [ [\frac{2}{3} * 1.5 ]](https://tex.z-dn.net/?f=%205.80%5E2%20%20%3D%20%203.00%5E2%20%2B%20%202%20%20%2A%20%20a%20%2A%20%20%5B%20%5B%5Cfrac%7B2%7D%7B3%7D%20%2A%201.5%20%5D)
=> ![5.80^2 = 3.00^2 + 2 * a * [1 ]](https://tex.z-dn.net/?f=%205.80%5E2%20%20%3D%20%203.00%5E2%20%2B%20%202%20%20%2A%20%20a%20%2A%20%20%5B1%20%5D)
=>
Generally magnitude of the force F during this interval in N is mathematically represented as
=> 
=>
Answer:
V_inside = 36 V
Explanation:
<u>Given </u>
We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.
<u>Required</u>
We are asked to calculate the potential at the centre of the sphere
<u>Solution</u>
The potential energy due to the sphere is given by equation
V = (1/4*π*∈o) × (q/r) (1)
Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V
V ∝ 1/r
The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next
V_1/V_2=r_2/r_1
V_inside/V_outside = r/R
V_inside = (r/R)*V_outside (2)
Now we can plug our values for r, R and V_outside into equation (2) to get V_inside
V_inside = (1.2 m )/(0.600)*18
= 36 V
V_inside = 36 V
F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
Explanation:
A sound wave is often called a pressure wave because there are regions of high and low pressure established in them medium through which the sound wave travels. The regions of high pressure are known as <u>Compressions</u> and the regions of low pressure are known as <u>Rarefactions</u> . Sound waves are composed of compressions and rarefactions. Compressions are the parts where the molecules are congusted and pressed together. However in the rarefactions molecules are relax and have enough space for expansion. Sound waves are the logitudnal waves and always been defined as the motion of the medium particles parallel to the wave motion.
