Answer:
A: The acceleration is 7.7 m/s up the inclined plane.
B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane
Explanation:
Let us work with variables and set
As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.
Part A:
From the free body diagram we see that the total force along the x-axis is:
Now the force of friction is 
where 
is the normal force and from the diagram it is 
Thus 
Therefore,
Substituting the value for 
we get:
Now acceleration is simply
The negative sign indicates that the acceleration is directed up the incline.
Part B:
Which can be rearranged to solve for t:
Substitute the value of 
and 
and we get:
which is our answer.
Notice that in using the formula to calculate time we used the positive value of 
, because for this formula absolute value is needed.
Answer:
Explanation:
It is given that,
Frequency for submarine communications, f = 76 Hz
We need to find the wavelength of those extremely low-frequency waves. the relation between the wavelength and the frequency is given by :
or
So, the wavelength of those extremely low-frequency waves is 4000 km. Hence, this is the required solution.
Answer:
C
Explanation:
From the question we are told that a vector on the x and y plane face their negative axis
Generally in the x and y plane thr negative y axis is made to face down opposite the positive y axis
Whilst the negative x axis faces the left which is also alternate to the positive x axis
Generally A vector pointing towards the x and y negative axis fro the origin (0) will definitely be in the third quadrant
<span>a. We can find the velocity when the camera hits the ground.
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera hits the ground with a velocity of 42 m/s
b. We can find the time it takes for the camera to hit the ground.
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
It takes 11.4 seconds for the camera to hit the ground.</span>
Answer:
The required radius is 2.62 cm
Explanation:
Given;
magnitude of current in wire, I = 1.31 A
magnetic field strength, B = 10 µT
Applying Biot Savart equation;
where;
r is the radius of the wire
μ₀ is constant = 4π x 10⁻⁷ Tm/A
I is the current in the wire
B is the magnetic field strength
Substitute the given values and calculate the radius
Therefore, the required radius is 2.62 cm
