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Nana76 [90]
3 years ago
10

you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point

Physics
1 answer:
tino4ka555 [31]3 years ago
4 0

No matter what direction you throw it, or with what speed, its acceleration is immediately 9.8 m/s^2 downward as soon as you release it from your hand, and it doesn't change until the ball hits something.

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A²  ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J  = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

3 0
3 years ago
Wo siblings are arguing over what way to pull a 10 kg wagon. Boris wants to pull it to the right. Boris puts 220 N of force on t
Mrrafil [7]

Answer:

a) Then net force acting on the wagon  30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

In y-axis:

The weight ( P = m*g =  10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force  Fn = P = 98 [N]  positive ( surface is frictionless)

∑Fy = 0     P - Fn = 0     P = Fn  = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N]  = 10 Kg * a                      [N] = Kg * m /s²

-30 [kg*m/s²]  = 10 * Kg * a

-30/10  =  - 3  m/s²  = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon  30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

5 0
3 years ago
Which will remain the same for two identical books, one lying flat and the other standing on an end?
Fiesta28 [93]

Weight will remain the same for two identical books, one lying flat and the other standing on an end.

The strain at a factor internal a liquid is at once proportional to the intensity of the factor. When an item is submerged in a liquid, the intensity of its backside from the floor of the liquid is extra than that of some other a part of the item.

Archimedes' precept is the declaration that the buoyant pressure on an item is identical to the load of the fluid displaced with the aid of using the item.

Ensure your scale is on a flat, strong and stage floor. Do now no longer use your scale on carpet. When taking measurements, stand nevertheless withinside the middle of the platform till all measurements are displayed, and if feasible do now no longer circulate your scale in-among measurements.

Learn more about weight here brainly.com/question/229459

#SPJ4

8 0
1 year ago
Does this graph represent constant or changing speed? How do you know? Find the average speed
stepan [7]
Constant speed because the time is directly proportional to the speed (2). The average speed is 2 m/s
6 0
3 years ago
The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete
sweet [91]

Answer:

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

F_n = 1.72 \times 10^4

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

v = \sqrt{5 R g}

also by energy conservation this is gained by initial potential energy

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

so we will have

\sqrt{2gh} = \sqrt{5Rg}

now we have

h = \frac{5R}{2}

here we have

R = 7.5 m

so we have

h = \frac{5(7.5)}{2}

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

now when it reach point C then the speed will be

mgh - mg(2R_c) = \frac{1}{2]mv_c^2

v_c^2 = 2g(h - 2R_c)

v_c = 13.1 m/s

now normal force at point C is given as

F_n = \frac{mv_c^2}{R_c} - mg

F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)

F_n = 1.72 \times 10^4

7 0
4 years ago
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