<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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Since the acceleration is uniform, we can calculate it from the data we are given:
a = (vf - vi)/2
where vf=33 m/s and vi=11 m/s
Then use Suvat's equation:
x(t) = vi*t + 0.5 * a * t
where t=20s
They are easy to use and more reliable
Answer:
C) Both technicians A and B
Explanation:
Fuel pressure regulators are a vacuum operated spring-loaded diaphragm that enables a vehicle's fuel delivery system to maintain a constant pressure. When the vehicle is at idle the regulator is open allowing fuel to bypass the delivery system and go back into the tank.
In pounds per square inch, the atmosphere exerts 14.7 PSI at sea level on average. The vacuum inside an engine's intake manifold, by comparison, can range from zero up to 22 inches Hg or more depending on operating conditions. Vacuum at idle is always high and typically ranges from 16 to 20 inches Hg in most vehicles.
Answer:
0.0102 m or 1 cm
Explanation:
Let g = 10m/s2
The potential energy of the shopping cart of the top of the hill is:
When the cart gets to the bottom of the hill, all this potential energy is converted to kinetic energy:
As the cart stop due to the stump, the can of peaches flies with the same speed.
By Newton's 3rd law, the car would exert a 490N force on the can too
The deceleration of the can would then be:
This force would stop the can, but not without making a dent, aka a traveled distance on the car skin
We can use the following equation of motion to find out the distance traveled by the can:
where v = 0 m/s is the final velocity of the can when it stops, 
= 40m/s is the initial velocity of the can when it hits, a = -1960 m/s2 is the deceleration of the can, and 
is the distance traveled, which we care looking for:
or 1 cm
