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2 years ago

ILL GIVE YOU ANYTHING TO ANSWER THIS!! IM HELPING MY CRUSH AND IDK THE ANSWER TO THIS PLZ HELP!

Physics

1 answer:

Luba_88 [7]2 years ago

7 0

C is your answer. PH is an acidic trait used to define just how acidic an acid is. Since you are classifying an acid it would go in C with PH.

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Each large pizza can feed three teenage boys. How many pizzas would be needed to feed the

Blababa [14]

Answer:

19 pizzas

Explanation:

1 pizza = 3 boys

=> 56 ÷ 3 = 18.666666666666.. or about 19 pizzas

6 0

3 years ago

Read 2 more answers

An object falls a distance h from rest. If it travels 0.460h in the last 1.00 s, find (a) the time and (b) the height of its fal

nadya68 [22]

Answer:

A. Time, t = 4.35s

B. Height of its fall, S = 92.72m

Explanation:

Vo = 0 m/s

Vi = 0.46h m/s

S = h m

a = 9.81 m/s2

To calculate the time taken, we need to get the value of the distance, h.

Using the equations of motion,

Vi^2 = Vo^2 + 2aS

Where Vi = final velocity

Vo = initial velocity

a = acceleration due to gravity

S = height of its fall

(0.46h)^2 = 0 + 2*9.81*h

0.2116h^2 = 19.62h

h = 19.62/0.1648

= 92.722 m

To calculate the time,

S = Vo*t +(1/2)*a*t^2

92.772 = 0 + (1/2)*9.81*t^2

t^2 = 185.44/9.81

= 18.904

t = sqrt(18.904)

= 4.348 s

7 0

3 years ago

Read 2 more answers

You wiggle a string, that is fixed to a wall at the other end, creating a sinusoidal wave with a frequency of 2.00 Hz and an amp

allochka39001 [22]

Answer:hhhkkzkxixuhhhdhhdhdhhhdhhhshsbbsbzhhdhhdhhhshhhdhhdusudhggdydyydhshdhgddhsjsudhdhhdh

Explanation:NzjzjhbxxhzhdghsjshshhHjajskajakakakakaiaiaiijhayayayagayahahah

4 0

3 years ago

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

3 years ago

The electric field strength E is measured as:

soldi70 [24.7K]

The correct answer is
<span>force per unit charge.

In fact, the electric field strength is defined as the electric force per unit charge experienced by a positive test charge located in the electric field. In formula:
</span> $E= \frac{F}{q}$
where
E is the electric field strength
F is the electric force experienced by the charge
q is the positive test charge.

4 0

3 years ago