Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.
Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s
Answer: 11.4 s
Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²
Answer: 44.7 m/s²
Answer: 5.71kg and 10.33N
Explanation:
Recall,
Weight =mass × g
Where w = 56.0 N , g = 9.8 m/s2
56= m × 9.8
M = 56/9.8
M = 5.71 kg
At the surface of the moon, the mass still remains 5.71kg because it is a fundamental quantity it does not change irrespective of position.
Weight at the moon surface will change as a result of change in gravity. Hence, weight = mg
Where g=1.81m/s2, m= 5.71
W = 5.71×1.81
W = 10.33N
Answer:
15 grams
Explanation:
9 sugar cubes + 2 teaspoons of baking soda + 2 grams of the sugar cubes each + 3 grams of baking soda = 15g.
Answer:
1440 W
Explanation:
The power drawn by a voltage source is the rate of energy consumed, and it can be written as:
where
V is the voltage
I is the current
P is the power
For the source in this problem we have:
I = 12 A is the current
V = 120 V is the voltage of the source
Solving for P, we find the power:
Answer:
9.67 A
Explanation:
The weight of a student with a mass of m = 75 kg is:
where g=9.8 m/s^2 is the acceleration due to gravity.
We want the magnetic force on the wire to be equal to this weight. The magnetic force on the wire is
where
I is the current in the wire
L = 2.0 m is the length of the wire
B = 38 T is the magnetic field
is the angle between the direction of B and L
Since we want W=F, we can write
And we can solve it to find the current I: