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jeka94
2 years ago
12

A juggler throws balls into the air. He throws one when the previous one is at its highest point. How high the balls rise if he

throws m balls per second?
Physics
1 answer:
VikaD [51]2 years ago
6 0
<span>If "m" balls are thrown per second, the time taken for a ball to reach its maximum height will be 1/m seconds. How to get this? See that the next ball is thrown only when the previous ball reaches its maximum height. If 'm' balls were thrown in 1 second this means that each ball was attaining its maximum ht in 1/m seconds. This was the main part. Now we can proceed to find maximum height in 2 ways- a) We know for upward journey ,
t=1/m
  a=-g
  v=u-gt
  final velocity ,v = 0 (at highest point) u
=gt = g/m
  Now we can apply h=ut-1/2 gt^2
  Putting the values of u,t, we will get h= g/2m^2
  b) The second method uses a trick that time taken to reach the maximum ht is same as time taken to fall down. So, we will now consider the downward journey of ball which also takes 1/m seconds We apply
 h=ut+1/2gt^2
Here u=0 ,t=1/m
We will again get ,
h=g/2m^2</span>
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A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate
frosja888 [35]

Answer:

N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons

Explanation:

The total charge is distributed over the two objects:

Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\

The plate and the rod must have Q_{total}/2\\. So the charge transferred from the plate to the rod is:

Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\

Number of electrons:

N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons

4 0
3 years ago
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
2 years ago
20 POINTS
Jobisdone [24]

Answer:

C) For every action, there is an equal and opposite reaction

Explanation:

Newton's Third Law

For every force, there is an equal and opposite reaction time. Explains why forces act in pairs.

Brainliest?

3 0
3 years ago
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
If the amplitude of the resultant wave is twice as great as the amplitude of either component wave, and the wave exhibits reinfo
Lunna [17]
Be in phase with each other! otherWise B
5 0
2 years ago
Read 2 more answers
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