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2 years ago

12

Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the tw

o charges such that the net force on Q0 will be zero?

1 answer:

azamat2 years ago

7 0

Answer:

yes independent of the sign or valve of Q

Explanation:

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A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s

inysia [295]

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from: It MUST have come from the hand that threw the ball. The hand would have needed to do <em>45.3 J</em> of work on the ball before releasing it.

6 0

2 years ago

A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many

jekas [21]

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

= 57units

5 0

2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago

A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 1

Blizzard [7]

Por definicion tenemos que
(F/A) = E(∆/0)
Sustituyendo los valores tenemos y despejando ∆:
∆ = (F/(πr2 × E))*0
(5000×5)/(3.14×(34×10^−2)^2×(125×10^8))
5.5×10^−6 m

4 0

1 year ago

Read 2 more answers

If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what

nataly862011 [7]

Answer:

The final temperature of the object will be 42.785 °C

Explanation:

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.

The equation for calculating the heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

Q= 450 J
c= 2.89 $\frac{J}{g*C}$
m= 20 g
ΔT= Tfinal - Tinitial= Tfinal - 35 °C

Replacing:

450 J= 2.89 $\frac{J}{g*C}$ *20 g* (Tfinal - 35°C)

Solving for Tfinal:

$\frac{450 J}{2.89\frac{J}{g*C}*20g} =Tfinal -35C$

7.785 °C=Tfinal - 35°C

7.785 °C + 35°C= Tfinal

42.785 °C=Tfinal

<u><em>The final temperature of the object will be 42.785 °C</em></u>

8 0

2 years ago

Read 2 more answers