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lions [1.4K]
3 years ago
5

A ball is dropped from the roof of a building. Points A, B, C, and D in the diagram below represent positions of the ball as it

falls. At which position will the ball have the greatest kinetic energy?

Physics
1 answer:
Anvisha [2.4K]3 years ago
4 0

Answer:

C. C

Explanation:

The ball will have its greatest kinetic energy just before it touches the ground which is at point C.

Kinetic energy of a body is the energy due to the motion of the body. As a body moves and its speed increases as a result of acceleration, then the kinetic energy will increase more.

As the body starts from rest, at point A, through B and C, the velocity increases drastically due the pull of the earth.

So, point C will show the greatest velocity and hence the most kinetic energy by the body.

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A block of mass 8 m can move without friction
nekit [7.7K]

Answer:

Let M1 = 8 kg and M2 = 34 kg

F = M a = (M1 + M2) a

F = M2 g     the net force accelerating the system

M2 g = (M1 + M2) a

a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2

5 0
2 years ago
Which of the following forms matter?<br> A. Proteins<br> B.atoms<br> C.cells<br> D. DNA
Artemon [7]

Answer:b) atoms

Explanation:which are in turn made up of protons, neutrons and electrons

8 0
3 years ago
Read 2 more answers
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?​
kykrilka [37]

Answer:

3A

Explanation:

Rtoal=R1+R2+R3=5+10+15=30

I=V/R 90/30

I=3

3 0
3 years ago
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
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