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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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Which of Earth's spheres have the greatest impact from the forest fire?

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I would say the atmosphere because air and oxygen are essential for human life however it is debatable. You should check out this prezi and judge for yourself which sphere is effected the most from forest fires based on the descriptions on how a forest fire effects all four spheres.

https://prezi.com/m/50ogw-qsyoxc/the-four-spheres/

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Having _________is the most obvious difference between a eukaryotic and a prokaryotic cell, but there are other differences as w

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Answer:

1. d. a nucleus

2. b. Membrane-bound organelles

3. c. ten time smaller

4. b. animal cell

5. c. Bacteria

Explanation:

Hope this helps... :)

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3 years ago

10 points) A rubber ball and a lump of putty of the same mass m = 0.05 kg are thrown with the same speed v=10.0 m/s in positive

snow_lady [41]

Answer:

the ball experiences the greater momentum change

Explanation:

You have to take into account that momentum change is given by

$\Delta p=mv_f-mv_b$

where vf and vb are the speed of the object after and before the impact.

In the case of the ball you have

$\Delta p=(0.05kg)(-10.0\frac{m}{s})-(0.05kg)(10.0\frac{m}{s})=-1kg\frac{m}{s}$

where the minus of vf is included due to the motion is in an opposite direction regarding with vb

And for the lump

$\Delta p=(0.05kg)(10.0\frac{m}{s})-(0.05kg)(0\frac{m}{s})=0.5kg\frac{m}{s}$

Hence, the ball experiences the greater change

hope this helps!!

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3 years ago

What does the term perihelion mean?

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3 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

3 years ago