Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time
Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>
Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration
Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N
<em>ANSWER: 240 N </em>
Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²
Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m
<em>ANSWER: d = 13.5 m</em>
On July 19th, 1985 Christa McAuliffe was selected as the first teacher to go to space by NASA. Her dream was to be the first teacher to go to space with all her lessons. Her daughter keeps her dream alive today by setting up 40 schools called the McAuliffe Centre.
<h3>Answer</h3>
6.6 N pointing to the right
<h3>Explanation</h3>
Given that,
two forces acting of magnitude 3.6N
angle between them = 48°
To find,
the third force that will cause the object to be in equilibrium
<h3>1)</h3>
Find the vertical and horizontal components of the two forces
vertical force1 = sin(24)(3.6)
vertical force2= -sin(24)(3.6)
<em>(negative sign since it is acting on opposite direction)</em>
vertical force3 = sin(24)(3.6) - sin(24)(3.6)
= 0
<h3>2)</h3>
horizontal force1 = cos(24)(3.6)
horizontal force2= cos(24)(3.6)
horizontal force3 = cos(24)(3.6) + cos(24)(3.6)
= 2(cos(24)(3.6))
= 6.5775 N
≈ 6.6 N
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Answer: 3A
Explanation:
First, we must calculate voltage by summing the reciprocal of the resistances in the circuit
1/40 +1/40 = 2/40 = 1/20 = 1/R, so R = 20Ω
Using Ohm's Law, I = V/R, so the current I is I = 120/20 = 6A
Using Kerchoff's Circuit Law, we know that current will split evenly at a junction, so each resistor will get one half, or 3A.
Answer:
Neither both are correct
Explanation:
The question involves the removal and installation procedure of cable for parking break. The answer for this could be found through the manual procedure for repairing parking break
Technician A is wrong because rear brakes should be adjusted before the parking cable adjustment is being made.
Technician B is wrong because the parking brake lever should be intact and secure at all clicks to allow maximum security. 15 clicks in this case should be the maximum number of clicks for the lever.
