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3 years ago

if a piece of sea floor has moved 50 km in 5 million years what is the yearly rate of sea-floor motion?

Physics

2 answers:

BartSMP [9]3 years ago

6 0

0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million

Sonbull [250]3 years ago

5 0

50 km = 50,000 meters

   (50,000 meters) / (5 million years)

   =    (5 x 10⁴) / (5 x 10⁶)    (meters/year)

   = 1 x 10⁻²    meter/year

   = 1 centimeter per year

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2. When the pump removed the air in the bell, the balloon contracted expanded

IRINA_888 [86]

When the pump removed the air in the bell, the balloon expanded.

Option: B

Explanation:

In order to construct our own environment in the glass jar known as bell jar system, which can be used to explore and consider our larger environment on Earths, for an instance. Here a glass jar that hinges on an airtight rubber basis i.e seals appropriately. At the top of the jar, a bung is connected to it which passed via a metal tube. It has an adjacent flexible tube that goes to a hand vacuum pump and the best hand-powered pump was made with a wine preserver.

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3 years ago

which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload

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2 years ago

Water (density = 1 ´ 103 kg/m3) flows at 10 m/s through a pipe with radius 0.030 m. the pipe goes up to the second floor of the

Hatshy [7]

density of water = $1000 kg/m^3$

velocity of flow = $10 m/s$

radius of pipe = $0.030 m$

Height of second floor = $2 m$

Now we can use here Bernuoli's Equation to find the speed of water flow at second floor

$P_1 + 1/2\rho v_1^2 + \rho g h_1= P_2 + 1/2 \rho v_2^2 + \rho g h_2$

$P + 1/2 * 1000 * 10^2 + 1000* 9.8 * 0 = P + 1/2 * 1000 * v^2 + 1000*9.8*2$

$v = 7.8 m/s$

Now in order to find the radius of pipe we can use equation of continuity

$A_1 v_1 = A_2 v_2$

$\pi *0.030^2 * 10 = \pi * r^2 * 7.8$

$r = 0.034 m$

So radius of pipe at second floor is 0.034 meter

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3 years ago

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

$W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m$

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

$W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J$

So, the new work is more than 130 J.

6 0

2 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

$W=K_f - K_i$

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

$K_f$ is the final kinetic energy