Most rocks contain more than one type of

All waves transmit energy. Only one type of wave does not require a medium to transmit energy. That is a _____wave?

Dmitrij [34]

Answer:

sound waves

Explanation:

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6 0

3 years ago

Read 2 more answers

I need help on this question!

Mashutka [201]

Explanation:

click the link you can get answer of the question

3 0

3 years ago

A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the

blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0

2 years ago

Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit such that

$P=VI=V^2/R$

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

$Current2=\frac{V}{2R}$

Power2=Voltage * Current2

$Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2$

$TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}$

6 0

2 years ago

Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains

lakkis [162]

Given that the density of heptane is

$d_h=\frac{0.684g}{mL}$

The mass of heptane is

$m_h=31\text{ g}$

The density of water is

$d_w=\frac{1g}{mL}$

The mass of water is

$m_w=37\text{ g}$

The volume of heptane will be

$\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}$

The volume of water will be

$\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}$

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

$\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}$

The total volume of liquid in the cylinder will be 82.32 mL.

7 0

8 months ago