Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
<u>Answer</u>: The potential difference across the resistor is 12 volts.
<u>Explanation:</u>
To calculate the potential difference cross the resistor, we use Ohm's Law. This law states that the potential difference across two wires is directly proportional to the current flowing through that wire.
Mathematically,
Where,
V = potential difference = ?V
I = Current flowing = 1.2 A
R = Resistor = 
Putting values in above equation, we get:
Hence, the potential difference across the resistor is 12 volts
The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
<span>If my memory serves me well, sensory receptors which would lead you to squint in bright light are called </span><span>C. photoreceptors</span>
Explanation:
F = 20N m= m1 a=10m/s²
m=m2 a=5m/s²
F = ma
<u>for the first one</u><u>:</u><u> </u>
f=m1 × a
20 = m1 ×10
20=10m1
m1=20/10
m1=2
<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>
f=m2×a
20=m2×5
m2= 20/5
m2= 4
since F=ma
F=(m1+m2) ×a
F =(4+2)×a
F =6×a
F=20(from the question above )
20=6×a
a=20/6
a=3.33
