Answer:
Workdone = 1960 Joules.
Explanation:
Given the following data;
Mass = 5kg
Force = 49N
Height (distance) = 40m
To find the workdone;
Workdone = force * distance
Substituting into the equation, we have;
Workdone = 49*40
Workdone = 1960 Joules.
Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.
A higher frequency. I just got this answer on usa test prep.
Answer:
B = 191.26 cm
θ = -14.73°
Explanation:
given,
magnitude of the first displacement(A) = 146 cm
at an angle of 124°
resultant magnitude = 137 cm
and angle made with x-axis by the resultant(R) = 32.0°
component of A in X and Y direction
A x = A cos θ = 146 cos 120° = -73 cm
A y = A sin θ = 146 sin 120° = 126.4 cm
now component of resultant in x and y direction
R x = 137 cos 35°
= 112.2 cm
R y = 137 sin 35°
= 78.6 cm
resultant is the sum of two vectors
R = A + B
R x = A x + B x
B x = 112.2 - (-73) = 185.2 cm
B y = R y - A y
B y = 78.6 - 126.4 = -47.8 cm
magnitude of B
B = 
B = 
B = 191.26 cm
angle
θ = -14.73°
I would make the ramp flatter. In doing so the ramp would have to be longer.
R = 0.407Ω.
The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.
To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.
We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:
R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]
R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²
R = 0.407Ω