Answer:
V = 3.17 m/s
Explanation:
Given
Mass of the professor m = 85.0 kg
Angle of the ramp θ = 30.0°
Length travelled L = 2.50 m
Force applied F = 600 N
Initial Speed u = 2.00 m/s
Solution
Work = Change in kinetic energy
The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
Answer:
i think its the third one or second one
Answer:
Potential difference = 6.0 V
I for 1.0Ω = 6 A
I for 2.0Ω = 3 A
I for 3.0Ω = 2 A
Explanation:
Potential difference (ΔV) = Current (I) x Resistance (R)
The potential difference is constant and equals 6.0 V, hence;
I = ΔV/R
When R = 1.0, I =6/1 = 6 amperes
When R = 2.0, I = 6/2 = 3 amperes
When R = 3.0, I = 6/3 = 2 amperes
<em>The potential difference is 6.0 V and the current is 6, 3, and 2 amperes for a resistance of 1.0, 2.0 and 3.0Ω respectively.</em>
Answer:
3 cm
Explanation:
According to the question,
.
.
.
Now the approximate slit's image width is equal to width of central maxima.
And width of central maxima is twice the width from center to first maxima
So,
.
Substitute all the variable in above equation.
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