Water on Earth was (a) transported here by comets; (b) accreted from the solar nebula; (c) produced by volcanoes in the form of steam; (d) created by chemical reactions involving hydrogen and oxygen shortly after Earth formed.

Answers

Answer 1

Answer:

Answer: Water on Earth was transported here by comets. The correct option is A.

Explanation:

Comets are made up of water with ice, rock and minerals.

Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.

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You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?

Answers

Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

$k=(2.303)/(t)\log(a)/(a-x)$

where,

k = rate constant = ?

t = time passed by the sample = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

$k=(2.303)/(90.3min)\log(400)/(100)$

$k=1.54* 10^(-2)\text{ min}^(-1)$

Now we have to calculate the half-life of substance, we use the formula :

$k=(0.693)/(t_(1/2))$

$1.54* 10^(-2)\text{ min}^(-1)=(0.693)/(t_(1/2))$

$t_(1/2)=45min$

Therefore, the half-life of this substance will be, 45 minutes.

A dynamics cart with a friction pad is placed at the top of an inclined track and released fromrest. The cart accelerates down the incline at a rate of 0.60 m/s2. If the track is angled at 10degrees above the horizontal, determine the coefficient of kinetic friction between the cart andthe track.

Answers

Answer:

0.114

Explanation:

There are two forces acting on the cart in the direction along the cart:

- The component of the gravitational force in the direction parallel to the ramp, $mg sin \theta$ , down along the ramp

- The force of friction, $\mu N$ , up along the ramp

So the equation of motion along this direction is:

$mg sin \theta - \mu N = ma$ (1)

where

m is the mass of the cart

$g=9.8 m/s^2$ is the acceleration due to gravity

$\theta=10^(\circ)$ is the angle of the ramp

$\mu$ is the coefficient of kinetic friction

N is the normal force exerted by the ramp on the cart

$a=0.60 m/s^2$ is the acceleration of the cart

The normal force can be found from the equation of the forces along the direction perpendicular to the ramp; in fact, the normal force is balanced by the component of the weight perpendicular to the ramp, so we have:

$N-mg cos \theta = 0$

From which we get:

$N=mg cos \theta$

Substituting into (1),

$mg sin \theta - \mu mg cos \theta = ma$

And solving for $\mu$ , we find the coefficient of friction:

$\mu = (g sin \theta - a)/(g cos \theta)=((9.8)(sin10^(\circ))-0.60)/((9.8)(cos 10^(\circ))=0.114$

What is the pH of a solution with a hydrogen ion concentration of 2.0x10^3.(Use 3 digits)

Answers

Answer:

2.70

Explanation:

pH = -log[H+]

pH = -log[2.0x10^-3]

pH = 2.70

The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 3.10 ✕ 10^−6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 43.0°C. When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 20.0 ✕ 10^10 N/m^2. (Assume the coefficient of thermal expansion of steel is 11 ✕ 10−6 (°C)−1.)

Answers

Answer:

361.46 N

Explanation:

$\alpha$ = Coefficient of thermal expansion = $11* 10^(-6)\ /^(\circ)C$

Y = Young's modulus for steel = $20* 10^(10)\ Pa$

A = Area = $3.1* 10^(-6)\ m^2$

$L_0$ = Original length = 1.28 km

$\Delta T$ = Change in temperature = $45-(-10)$

Length contraction is given by

$\Delta L=\alpha L_0\Delta T$

Also,

$\Delta L=(L_0T)/(YA)$

$\alpha L_0\Delta T=(L_0T)/(YA)\n\Rightarrow T=\alpha \Delta TYA\n\Rightarrow T=11* 10^(-6)* (43-(-10))* 20* 10^(10) * 3.1* 10^(-6)\n\Rightarrow T=361.46\ N$

The tension in the wire is 361.46 N

What is the answer to life and everything in the universe?

Answers

The answer to life and the universe is 42

What problem did Katherine face in checking other people’s calculations in the movie ¨Hidden figures¨

Answers

Answer:

Information from the leading mathematicians was considered "classified".

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