Assuming the power delivered by the horse does not change, the speed of the cart will decrease.
In fact, the power delivered by the horse is the work done by the horse (W) per unit time (t):
<span>If several bags are added to the cart, the horse must do more work to transport them. Therefore, W in the fraction increases. But if the power P of the horse is constant, then it means that the time t must increase as well. So, the horse will take more time to transport the car, and this means that the speed of the cart has decreased.</span>
Answer:
If the force applied is larger than 185.2 N, yes.
Explanation:
In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:
where
is the coefficient of static friction
is the mass of the table
is the gravitational acceleration
Substituting,
So, we are able to move the table if we push with a force larger than 185.2 N.
You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.
First, convert 60 mph to m/s:
60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s
Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.
free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)
H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m
If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.
Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.
Answer:
a) 588,000 N
b) 294000 N
Explanation:
Given that
Density of water = 1000kg/m3
(g) = 9.8m/s2
volume is given as (V)= 5m*4m*3m
a) force will be equal to weight of water
b) at either end
[A = wh]
F = 294000 N
Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
