Which of these units is most appropriate to use when measuring the mass of a person?

A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the elect

FromTheMoon [43]

Complete question:

A uniform electric field is created by two parallel plates separated by a

distance of 0.04 m. What is the magnitude of the electric field established

between the plates if the potential of the first plate is +40V and the second

one is -40V?

Answer:

The magnitude of the electric field established between the plates is 2,000 V/m

Explanation:

Given;

distance between two parallel plates, d = 0.04 m

potential between first and second plate, = +40V and -40V respectively

The magnitude of the electric field established between the plates is calculated as;

E = ΔV / d

where;

ΔV is change in potential between two parallel plates;

d is the distance between the plates

ΔV = V₁ -V₂

ΔV = 40 - (-40)

ΔV = 40 + 40

ΔV = 80 V

E = ΔV / d

E = 80 / 0.04

E = 2,000 V/m

Therefore, the magnitude of the electric field established between the plates is 2,000 V/m

7 0

2 years ago

A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo

aliina [53]

For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative.
So Vf = square root [2(-9.8)(11.0) + (18.0)^2] 
Vf = 10.4 m/s your answer is correct.

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, 11= 18t m/s-(1/2) 9.8t^2 then -11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s

5 0

3 years ago

If jack was traveling north 120 miles and it took him 3 hours to get there. what is the velocity that jack was traveling? is thi

Dimas [21]

Answers:

40 mp/h; Vector

Reason:

120/3 is 40 miles per hour.

Velocity is a vector measurement.

^.^

- Amanda

4 0

3 years ago

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago

Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea

Alexxx [7]

Answer:

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

Em₀ = U = m g h

Final point. To get off the ramp

Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

v = w r

w = v / r

we substitute

mg h = ½ v² (m + I / r²)

v² = 2 gh $\frac{m}{m+ \frac{I}{r^2} }$

v² = 2gh $\frac{1}{1 + \frac{I}{m r^2} }$

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

v² = v₀² + 2 a L

where L is the length of the plane

v² = 2 a L

a = v² / 2L

we substitute

a = $g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }$

let's use trigonometry

sin θ = h / L

we substitute

a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

I = m r²

we look for the acerleracion

a₁ = g sin θ $\frac{1}{1 + \frac{mr^2 }{m r^2 } }$ 1/1 + mr² / mr² =

a₁ = g sin θ ½

b) solid cylinder

I = ½ m r²

a₂ = g sin θ $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ $\frac{1}{1+ \frac{1}{2} }$

a₂ = g sin θ ⅔

c) hollow sphere

I = 2/3 m r²

a₃ = g sin θ $\frac{1}{1 + \frac{2}{3} }$

a₃ = g sin θ $\frac{3}{5}$

d) solid sphere

I = 2/5 m r²

a₄ = g sin θ $\frac{1 }{1 + \frac{2}{5} }$

a₄ = g sin θ $\frac{5}{7}$

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ $\frac{105}{210}$

b) a₂ = g sinθ ⅔ = g sin θ $\frac{140}{210}$

c) a₃ = g sin θ $\frac{3}{5}$ = g sin θ $\frac{126}{210}$

d) a₄ = g sin θ $\frac{5}{7}$ = g sin θ $\frac{150}{210}$

the order of acceleration from lower to higher is

a₁ <a₃ <a₂ <a₄

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0

3 years ago