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3 years ago

The difference between the Carbon 12 and Carbon 14 isotopes is the

1 answer:

4 0

Answer:
The number of neutrons.

Explaination:
The atoms of carbon 12 contain 6 neutrons while the atoms of carbon 14 contain 8 neutrons

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While John is traveling along an interstate highway, he notices a 150 mi marker as he passes through town. Later John passes a 1

Lerok [7]

Answer:

Part a)

distance = 112 miles

Part b)

current position = 112 miles from the position of town

Explanation:

Part a)

Since the distance marker is showing the distance between the town and the position of john at all time

so here we have

$d = 112 miles$

Part b)

Current position of John is given as

$r = 112 miles$

from the position of the town

7 0

4 years ago

Read 2 more answers

What does this picture mean?

ivann1987 [24]

be aware of large trucks on the roadway
Tip: usually, these signs are hazard signs or warning.
Also, this isnt physics

5 0

3 years ago

Read 2 more answers

Num determinado equipamento industrial, um líquido de calor específico 0,50 cal/g.°C, entra a 20 °C e sai a 80 °C. Se a vazão de

balu736 [363]

Answer:

Tonta nadie sabe eso

Explanation:

4 0

3 years ago

What is he total resistance? A. 20 ohms B. 90 ohms C. 40 ohms D. 30 ohms

ehidna [41]

Imma guess A! Idk if it’s 100% correct tho so I’d check that!

6 0

3 years ago

Read 2 more answers

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

3 years ago