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vlabodo [156]
4 years ago
14

An 8.30 kg crate is pushed with a 17.7 N force. How fast does it accelerate?

Physics
2 answers:
kramer4 years ago
4 0

Answer:

2.13m/s^2

Explanation:

Use equation:

Fnet=ma

If 17.7 is the netto force (Fnet) then you just substitute the values in and solve.

Note m is mass (kg) and a is acceleration (m/s^2)

katovenus [111]4 years ago
4 0
F=17.7N
m=8.30kg
a=?
F=ma
a=F/m
a=17.7/8.30 (m/s^2)
I’m too lazy to use a calculator so yea. Just divide 17.7 by 8.30 and add unites (m/s^2)
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A semicircular plate with radius 6 m is submerged vertically in water so that the top is 2 m above the surface. Express the hydr
dalvyx [7]

Answer: 313920

Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.

Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.

The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.

The depth below the water surface of each strip is,

di = 8 − yi

and that in turn gives us the pressure on the strip,

Pi =ρgdi = 9810 (8−yi)

The area of each strip is,

Ai = 2√4− (yi) 2Δy

The hydrostatic force on each strip is,

Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy

The total force on the plate is found on the attached image.

5 0
3 years ago
Pls I need help for the problem solving part.
Bogdan [553]

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3 0
3 years ago
Which vector goes from (-1,-2) to (3, 4)?<br><br> A. d<br> B. c<br> C. b<br> D. a
kogti [31]

vector b is the answer...............

5 0
3 years ago
Two solid marbles A and B with a mass of 3.00 kg and 6.50 kg respectively have an elastic collision in one dimension. Before col
Akimi4 [234]

Answer:

va = 4.79 m/s

vb = 1.29 m/s

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂

22.75 = 3v₁ + 6.5v₂

For an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²

79.625 = 3v₁² + 6.5v₂²

Two equations, two variables.  Solve with substitution:

22.75 = 3v₁ + 6.5v₂

22.75 − 3v₁ = 6.5v₂

v₂ = (22.75 − 3v₁) / 6.5

79.625 = 3v₁² + 6.5v₂²

79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²

79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5

517.5625 = 19.5v₁² + (22.75 − 3v₁)²

517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²

0 = 28.5v₁² − 136.5v₁

0 = v₁ (28.5v₁ − 136.5)

v₁ = 0 or 4.79

We know v₁ isn't 0, so v₁ = 4.79 m/s.

Solving for v₂:

v₂ = (22.75 − 3v₁) / 6.5

v₂ = 1.29 m/s

8 0
3 years ago
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viva [34]

Answer:

a) g.p.e.=mass × gravitational field strength × height

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9800 (J)

4 0
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