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Drupady [299]
3 years ago
5

The ancient fish fossils were dated using C-14. If 1/16 the original amount of C-14 remains in the skeletons, approximately how

old are the fish. The half-life of C-14 is 5730.
a) 11,460 years
b) 17,190 years
c) 22,920 years
d) 28,650 years​
Physics
1 answer:
DIA [1.3K]3 years ago
5 0

Answer:

c) 22,920 years

Explanation:

The half-life of C-14 is 5730 so in 5730 years , amount of C-14 is reduced to half.

In 5730 x 2  years , the amount of C-14 is reduced to 1/4 .

In 5730 x 3 years ,  the amount of C-14 is reduced to 1/8 .

In 5730 x 4 years ,  the amount of C-14 is reduced to 1/16 .

In 22920 years ,  the amount of C-14 is reduced to 1/16 .

So the answer is 22920 years.

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How many work is done when a force of 33n pulls wagon 13meters
goldenfox [79]
Work = force x distance
13 \times 33 = 429
the answer is 429 joules



good luck
3 0
3 years ago
(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface
Keith_Richards [23]

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}

The expression of escape velocity is given by

v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s

The escape speed is 1777.92 m/s

3 0
3 years ago
Read 2 more answers
A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea
Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
R_T = R_1 + R_2 + ... + R_n

In parallel:
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

V = iR\\\\V = (6)(15) = \boxed{90 V}

4 0
2 years ago
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

where E = electric field intensity

5 0
3 years ago
Mechanical energy is conserved in the presence of which of the following types of forces?magnetic
Tju [1.3M]
<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.  

In this sense, the following forces are conservative:  

-Gravitational  

-Elastic

-Electrostatics  

While the Friction Force and the Magnetic Force are not conservative.

According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

7 0
3 years ago
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