A ball is dropped from a 20.0 m high tower.

a. How long will it take the ball to reach the ground?

b. What is the velocity of the ball just before it hits the ground?

Answers

Answer 1

Hi there!

We can begin by using the derived kinematic equation:

[tex]t = \sqrt{\frac{2h}{g}[/tex]

Plug in the given values and let g = 9.8 m/s²:

[tex]t = \sqrt{\frac{2(20)}{g}} = \boxed{2.02 s}[/tex]

Now, we can solve for its final velocity using the equation:

[tex]v_f = v_i + at[/tex]

It is dropped from rest, so vi = 0 m/s.

[tex]v_f = at[/tex]

[tex]v_f = 9.8(2.02) = \boxed{19.796 m/s}[/tex]

Answer 2

Answer:

a. The ball would reach the ground in approximately [tex]2.02\; \rm s[/tex].

b. The velocity of the ball right before landing would be approximately [tex]19.8\; \rm m\cdot s^{-1}[/tex].

(Assumptions: the ball was dropped with no initial velocity; air resistance on the ball is negligible; [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].)

Explanation:

Under these assumptions, the acceleration of this ball would be constantly [tex]a = g = 9.81\; \rm m\cdot s^{-2}[/tex] (same as the gravitational field strength) during the descent.

Displacement of the ball: [tex]x = 20.0\; \rm m[/tex].

Initial velocity of the ball: [tex]v_{0} = 0\; \rm m\cdot s^{-1}[/tex].

Let [tex]t[/tex] denote the duration of this descent.

The SUVAT equation [tex]x = (1/2)\, a\, t^{2} + v_{0}\, t[/tex] relates the known quantities [tex]a[/tex], [tex]x[/tex], and [tex]v_{0}[/tex] to the unknown [tex]t[/tex].

Substitute the known quantities into this equation and solve to find the value of [tex]t\![/tex]:

[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2} + 0\; {\rm m \cdot s^{-1}} \times t[/tex].

[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2}[/tex].

[tex]\displaystyle t^{2} = \frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}[/tex].

Since [tex]t > 0[/tex]:

[tex]\begin{aligned}t &= \sqrt{\frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}} \\ &\approx 2.01928\; \rm s\\ &\approx 2.02\; \rm s \\ & (\text{Rounded to 2 sig. fig.})\end{aligned}[/tex].

Since the acceleration of this ball is constant, the velocity of the ball right before landing would be:

[tex]\begin{aligned}v_{1} &= a\, t \\ &\approx 9.81\; \rm m\cdot s^{-2} \times 2.01928\; \rm s \\ &\approx 19.8\; \rm m \cdot s^{-1} \end{aligned}[/tex].


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Hi there!

Recall the conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'

Let m1 = 5 kg ball and m2 = 10 kg ball

Since m2 is at rest, we can rewrite:

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m1 =5kgm2=10( since it's at rest)v1=2m/sv2=0(since it's at rest)v1' = -1To find ↷the speed of the larger ball after the collisionSolution↷

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I would like to know if it would be m/s² or just m/s and explain why. thanks​

Answers

Solution:

It will be

[tex] {m/s}^{2} [/tex]

I am just explaining with the units only, because you want to know the accurate unit.

[tex] \frac{m/s}{s} \\ = \frac{m}{s} \times \frac{1}{s} \\ = \frac{m}{s \times s} \\ = \frac{m}{ {s}^{2} } \\ = {m/s}^{2} [/tex]

Hope you understood.

Do comment if you have any query.

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The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 1100 kg

acceleration = 0.5 m/s²

We have

force = 1100 × 0.5 = 550

We have the final answer as

550 N

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Answer is attached. I do not know if you needed to use 9.81 or 10 for the acceleration due to gravity, so both solutions are attached.

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3) both moving and non moving objects

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Answers

Considering the Law of Universal Gravitation, the mass of the second object is 78090.19 kg.

The Law of Universal Gravitation is a law that states that bodies, by the simple fact of having mass, experience a force of attraction towards other bodies with mass, called gravitational force.

This law establishes that every particle attracts any other particle with a force directly proportional to the product of the masses of both and inversely proportional to the square of the distance that separates them:

[tex]F=G\frac{m1m2}{r^{2} }[/tex]

where m1 and m2 are their masses; r the distance between them and G a universal constant that is called the constant of gravitation.

In this case, you know:

F= 0.128 NG=6.67× 10⁻¹¹ Nm²/kg²m1= 1300 kgm2= ?r= 0.23 m

Replacing:

[tex]0.128 N=6.67x10^{-11}\frac{Nm^{2} }{kg^{2} } \frac{1300 kg m2}{(0.23 m)^{2} }[/tex]

Solving:

[tex]0.128 N=1.639x10^{-6}\frac{N}{kg }x m2[/tex]

m2=0.128 N ÷ 1.639×10⁻⁶ [tex]\frac{N}{kg}[/tex]

m2=78090.19 kg

In summary, the mass of the second object is 78090.19 kg.

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A crate of mass 50kg is pulled up a rough inclined plane, inclined at an angle of 60° to the horizontal. Determine the total work done in moving the crate to the top of the inclined 200m high.​

Answers

Answer:

Wc = m*g = 50kg * 9.8N/kg = 490 N = Wt.

of crate.

Fc = 490N  60o = Force of crate.

Fp = 490*sin60 = 424.4 N. = Force

parallel to incline.

Fv = 490*cos60 = 245 N. = Force perpendicular to the incline.

Ff = u*Fv = u*245 = Force of friction.

Fap-Fp-Ff = m*a

Fap-424.4-u*245 = m*0 = 0

Fap = 424.4 + 245u. = Force applied.

L = 200m/sin60 = 231 m. = Length of

incline.

Work = Fap * L=(424.4+245u) * 231.

Explanation:

6. Explain a change that you can make in your diet to further your overall health.

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Answer:

“Eat more delicious whole-food plants at every meal and snack. Fruits, vegetables, whole grains, nuts and seeds are packed with nutrients and contain satisfying fiber that is good for digestion, disease prevention and sustained energy.

Explanation:

Make sure you are following the recommended amount of nutrients needed in your diet (depending on age and sex). You need enough carbohydrates (for energy) but also some fats. 40% of your diet should consist of fruits and vegetables, 25% of fibre rich carbohydrates, 25% of proteins and 10% fats.

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