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3 years ago

8

Using the equation for for Newton's Second law, m=F/a solve the following problem. You have been given an object with a force of

10N and an acceleration of 2 m/s2, what is the mass?
Group of answer choices

1. 8g

2. 3g

3. 20g

4. 5g

1 answer:

inessss [21]3 years ago

8 0

Answer:

4. 5g

Explanation:

F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.

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love history [14]

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3 years ago

Is there carbon atoms in the compound Ca3N2

katen-ka-za [31]

No, there are not any carbon atoms inside this compound.

The compound is $Ca_3N_2$ . This means there are 3 "Ca" atoms and 2 "N" atoms.

Ca is calcium
N is nitrogen

Thus, none of the elements in this compound are carbon, meaning there are no carbon atoms. Let me know if you need any clarifications, thanks!

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3 years ago

Read 2 more answers

A $100 cart is moving down an inclined plane at 1.30 m/s. The cart collides with a stationary object and rebounds with a speed o

Rasek [7]

Answer:

0.65 kg*m/s and 0.165 kg*m/s

Explanation:

Step one:

given data

mass m= 0.5kg

initial velolcity u=1.3m/s

final velocity v= 0.97m/s

Required

The change in momentum

Step two:

We know that the expression for impulse is given as

Ft= mv

Ft= 0.5*1.3

Ft= 0.65 kg*m/s

The expression for the change in momentum is given as

P= mΔv

substitute

Pt= 0.5*(1.3-0.97)

Pt= 0.5*0.33

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2 years ago

Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the

Alex777 [14]

The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire.

The magnetic field generated by wire 1 is:
$B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}= 4 \cdot 10^{-5}T$

The magnetic field generated by wire 2 is:
$B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T$

And so, the resultant magnetic field at the point midway between the two wires is
$B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T$

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3 years ago

Suppose the roller coaster in Fig. 6-38 (h1 = 33 m, h2 = 13 m, h3 = 25) passes point A with a speed of 2.10 m/s. If the average

Kamila [148]

Answer:

22.67 m/s

Explanation:

See in the picture.

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3 years ago