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3 years ago

Only 5 question plz answer

Physics

2 answers:

Aleks04 [339]3 years ago

4 0

It is subduction;) good luck on the others my man

icang [17]3 years ago

4 0

For question 7 it is the upper portion of the mantle and the crust

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7. If 8 million kg of water flows over Niagara Falls each second, calculate the power available at the bottom of the falls.

Alexxx [7]

Answer:

The power will be "3.92×10⁹ Watts". A further explanation is given below.

Explanation:

The given values as per the question,

Rate,

= 8 million kg

Distance,

= 50 m

Gravity,

= 9.8 m/s²

As we know,

The power will be:

⇒ $Power = Rate\times Distance\times Gravity$

On putting the values, we get

⇒ $= 8\times 10^6\times 50\times 9.8$

⇒ $=3.92\times 10^9 \ Watts$

7 0

3 years ago

What happens to beaches over time?

Rama09 [41]

the answer is d i took the test be cause i go to a k12 onlie school

7 0

3 years ago

Read 2 more answers

ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of

sesenic [268]

Answer:

a) $V_{2/1}=0.8m/s$

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

$V_{2/1} = V_{2} - V_{1} = 0.8m/s$

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

$t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s$

$t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s$

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

6 0

3 years ago

Read 2 more answers

A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t

sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, $\rho=2.44\times 10^{-8}\ \Omega-m$

Resistance in terms of resistivity is given by :

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

Also, V = IR

So,

$\dfrac{V}{I}=\dfrac{\rho l}{A}$

A is area of wire,

$\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}$ , r is radius, r = d/2 (diameter=d)

$\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m$

Out of four option, near option is (C) 17 μm.

6 0

3 years ago

Factor the expression 27x-18y

Eva8 [605]

Answer:

9(3x-2y)

Explanation:

27x-18y, the common factor here is 9

9(3x-2y)

7 0

3 years ago