The answer would be Gravity.
Gravity is pulling the weight down, which is pulling the car up the ramp.
Here's what you need to memorize for your exam tomorrow.
Distance = (speed) x (time) .
That's it. Memorize it.
-- If the question wants you to find speed, use it exactly in that form.
-- If the question wants you to find speed, then divide each side by (time)
and it says
. Time = (distance) / (speed) .
-- If the question wants you to find time, then divide each side by speed,
and it says
. Time = (distance) / (speed) .
So if you memorize that one equation ... Distance = (speed) x (time) ...
you can solve ANY problem to find distance, speed, or time.
On the sheet in the picture . . . . .
#2). The question is "How long ?". That's TIME that you have to find.
Use the equation in the form of
. TIME = (distance) / (speed)
. = (60 km) / (48 km/h)
. = 1.25 hours .
#3). This one wants you to find SPEED. Use the equation in the form of
. SPEED = (distance) / (time)
but be careful. The time has to be in hours. 55 minutes = 55/60 of an hour.
. SPEED = (distance) / (time)
. = (60 km) / (55/60 hour)
. = (60 x 60 km) / (55 hour)
. = 65.45 km/hr .
#4). This one wants you to find TIME. (It says "How long ?".)
It's two trips, so you have to find the time for each trip.
First trip: TIME = (distance)/(speed) = (24 km)/(65 km/hr) = 0.369 hr
Second trip: TIME = (distance)/(speed) = (50 km)/(80 km/hr) = 0.625 hr
Total time for both trips = (0.369 hr) + (0.625 hr) = 0.994 hour
<span>s=2.7 centimeters = 0.027 meters
t=3.9 milliseconds = 0.0039 seconds
s=(1/2)a*t^2
so
a=(2.7*2)/(0.0039)^2
= 355,029.58 m/s^2
a=355,029.58 m/s^2 = 355.02958 km / s^2</span>
Our sun is a medium mass star, so it wouldn't be too different from the sun's life cycle. It is born, lives for about 10 billion years and then dies. ... As a medium mass star nears the end of its life, it runs out of hydrogen which it has been fusing onto helium in its core for its whole life.
(a) The stress in the post is 1,568,000 N/m²
(b) The strain in the post is 7.61 x 10⁻⁶
(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.
<h3>Area of the steel post</h3>
A = πd²/4
where;
d is the diameter
A = π(0.25²)/4 = 0.05 m²
<h3>Stress on the steel post</h3>
σ = F/A
σ = mg/A
where;
- m is mass supported by the steel
- g is acceleration due to gravity
- A is the area of the steel post
σ = (8000 x 9.8)/(0.05)
σ = 1,568,000 N/m²
<h3>Strain of the post</h3>
E = stress / strain
where;
- E is Young's modulus of steel = 206 Gpa
strain = stress/E
strain = (1,568,000) / (206 x 10⁹)
strain = 7.61 x 10⁻⁶
<h3>Change in length of the steel post</h3>
strain = ΔL/L
where;
- ΔL is change in length
- L is original length
ΔL = 7.61 x 10⁻⁶ x 2.5
ΔL = 1.9 x 10⁻⁵ m
Learn more about Young's modulus of steel here: brainly.com/question/14772333
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