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MrRa [10]
3 years ago
10

Which of the following best describes a plane?

Physics
2 answers:
katrin [286]3 years ago
4 0
B, the surface of a flat table.
DerKrebs [107]3 years ago
3 0
Yes B, The surface of a flat table
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Can an object have increasing speed while its acceleration is decreasing?
alexdok [17]
The best option is C. This is due to friction.
6 0
3 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
3 years ago
A soccer player kicks a soccer ball at +10.0 m/s at an angle of 60°. What are the horizontal and vertical components of velocity
Nastasia [14]
This seems like a calculus problem. I'm assuming you would use cos and sin. so here's the vertical component +10.0m/s multiplied by sin60 = 8.66 rounded to the hundreths place. Now for horizontal, that would be +10.0m/s multiplied by cos60 = 5. hope this helped.
3 0
3 years ago
Read 2 more answers
In a tank full of water, the pressure on a surface 2 meters below the water level is 1.5 kPA. What's the pressure on a surface 6
Lina20 [59]

Answer:

P = 40.7kPa

Explanation:

To find the pressure on a surface 6 meter below you use the following formula, which takes into account the heights in which pressures are measured and also the density of the fluid and the gravitational acceleration:

P_2-P_1=-\rho g(y_2-y_1)             (1)

P2: pressure for a height of -6 m = ?

P1: pressure for a height of -2 m = 1.5kPa = 1500 Pa

ρ: density of water = 1000kg/m^3

g: gravitational acceleration = 9.8 ms^2

y2: -6m

y1: -2m

(the height is measure from the water level, because of that, the heights are negative)

You solve the equation (1) for P1:

P_1=P_2-\rho g(y_2-y_1)         (2)

Next, you replace the values of all variables in equation (2):

P_2=1500Pa-(1000kg/m^3)(9.8m/s^2)(-6-(-2))m=40700Pa\\\\P_2=40.7kPa

hence, the pressure on a surface 6 m below the water level is 40.7kPa

5 0
3 years ago
What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.
soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0
3 years ago
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