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lisabon 2012 [21]
2 years ago
5

Why do we use days and hours instead of hours?​

Physics
1 answer:
MA_775_DIABLO [31]2 years ago
4 0

Answer:

becuase counting by hours is more complicated. then days, to make it easier to understand instead of saying I will see you in 48 hours... its easier to understand and think to say see you in 2 days!

Explanation:

hope this helps. also we did not determine time. people from history did so... wasnt really my choice or anyones!

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When does a ball that is thrown vertically upwards reach its maximum speed?
ladessa [460]
As soon as you let go of it it is at its max speed because gravity is constantly pulling it down
4 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h
Neko [114]

Answer

given,

firm is producing  = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

\dfrac{MP_l}{MP_c}=\dfrac{40}{25}

\dfrac{MP_l}{MP_c}=\dfrac{8}{5}

\dfrac{W}{R}=\dfrac{20}{10}

\dfrac{W}{R}=2

\dfrac{MP_l}{MP_c} < \dfrac{W}{R}

Firm is not minimizing the cost because the firm use more capital and less labor.

3 0
3 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\&#10;mg=m\frac{v^2}{r}\\&#10;gr=v^2\\&#10;v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
Zach, whose mass is 90 kg , is in an elevator descending at 12 m/s . the elevator takes 3.0 s to brake to a stop at the first fl
sweet-ann [11.9K]
Finding out the acceleration 12/3 = 4m/s^2
thus it is descending so the actual acceleration would be 9.8-4 = 5.8 m/s^2
the weight will be 90*5.8 = 522 N
522/9.8 = 53.2 kg
7 0
3 years ago
Read 2 more answers
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