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Pie
3 years ago
12

4. O/N 15/P11/Q11

Physics
1 answer:
anygoal [31]3 years ago
5 0

Answer:

B) 4500 Pa

Explanation:

As pressure is force per unit area,

P = F/A

It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.

The possible areas are

0.30(0.40) = 0.12 m²

0.30(0.50) = 0.15 m²

0.40(0.50) = 0.20 m²

The pressure when the face with the largest area (0.20 m²) is down is

P = 900 / 0.20 = 4500 N/m² or 4500 Pa

the other possible pressures would be

900/0.15 = 6000 Pa

900/0.12 = 7500 Pa

which are both larger than our solution.

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The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which
emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

=  .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s  

8 0
3 years ago
A 5 kg
kotykmax [81]

Answer:

<em>The object-Earth system is open</em>

F_n=ma=5\ (7.2)=36\ N

Explanation:

<u>Accelerated Motion </u>

When an object is released in free air (with no other forces than the gravity), it describes a free-fall motion and the formulas include the acceleration of gravity as part of the calculations. But when there is another external force, then the acceleration is not the gravity, but the result of the net force exerted on the mass of the object.

By definition, an open system includes the exchange of energy from and to the surroundings, that is why all systems surrounding our planet are considered as open systems. In our case, the object is interacting with the planet's gravity and there is some other external force, which will be computed later. The object-Earth system is open.

If the object starts from rest, its initial speed is zero, and

v_f=a\ t

where a is the acceleration and t is the time. The distance traveled is given by :

\displaystyle y=\frac{a\ t^2}{2}

From the two above equations, we find that:

v_f^2=2ay

Solving for a

\displaystyle a=\frac{v_f^2}{2y}

\displaystyle a=\frac{12^2}{2\ (10)}

a=7.2\ m/s^2

It means the net force is

F_n=ma=5\ (7.2)=36\ N

The object's weight is

W=5\ (9.8)=49 N

This means there is some external force acting upwards delaying the object's fall of a magnitude of

F_e=49-36=13\ N

3 0
3 years ago
help me solve this: a bicyclist in the tour de france crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farth
Lana71 [14]
First we have to convert:
75 km / h ( * 3.6 ) = 270 m/s
18 m/s = 64.8 m/s
d = v i · t + 1/2 a · t²
v = v i + a t
-----------------------------------
4000 = 64.8 · t + 1/2 a · t²
270 = 64.8 + a t
a = ( 270 - 64.8 ) / t
a = 205.2 / t
4000 = 64.8 t + 1/2 · ( 205.2 / t ) · t²
4000 = 64.8 t + 102.6 t
4000 = 167.4 t
t = 4000 : 167.4 = 23.89 s
a = 205.2 : 23.89 = 8.58 m/s²
Answer: His average acceleration was 8.58 m/s²

7 0
3 years ago
How can you tell that something is in motion?
Westkost [7]
Answer - An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.
3 0
3 years ago
Read 2 more answers
During a baseball game, a batter hits a popup to a fielder a distance d m away. the acceleration of gravity is 9.8 m/s 2 . if th
s344n2d4d5 [400]

solution:

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3 0
3 years ago
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