Explanation:
Momentum is mass times speed.
p = mv
a) p = (1500 kg) (25.0 m/s) = 37,500 kg m/s
b) p = (40,000 kg) (1.00 m/s) = 40,000 kg m/s
The truck has more linear momentum.
Momentum in the y direction:
pᵧ = (1500 kg) (25.0 m/s) = 37,500 kg m/s
Momentum in the x direction:
pₓ = (1500 kg) (15.0 m/s) = 22,500 kg m/s
Total linear momentum:
p² = pₓ² + pᵧ²
p² = (22,500 kg m/s)² + (37,500 kg m/s)²
p = 43,700 kg m/s
Speed is constant. 50 miles = 1 hour. 600/50 = 12. 1hr(12) = 12 hours.
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
Answer:
B
Explanation:
V=IR I= curren V=volt R=resistor
8=2.R 8/2=R R=4
<h3>
Answer:</h3>
225 meters
<h3>
Explanation:</h3>
Acceleration is the rate of change in velocity of an object in motion.
In our case we are given;
Acceleration, a = 2.0 m/s²
Time, t = 15 s
We are required to find the length of the slope;
Assuming the student started at rest, then the initial velocity, V₀ is Zero.
<h3>Step 1: Calculate the final velocity, Vf</h3>
Using the equation of linear motion;
Vf = V₀ + at
Therefore;
Vf = 0 + (2 × 15)
= 30 m/s
Thus, the final velocity of the student is 30 m/s
<h3>Step 2: Calculate the length (displacement) of the slope </h3>
Using the other equation of linear motion;
S = 0.5 at + V₀t
We can calculate the length, S of the slope
That is;
S = (0.5 × 2 × 15² ) - (0 × 15)
= 225 m
Therefore, the length of the slope is 225 m
