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Zielflug [23.3K]
2 years ago
14

QUICKLY I AM IN THE LESSON NOWWW

Physics
1 answer:
vladimir2022 [97]2 years ago
7 0

Answer:

I don't really know much about the whole stars thing, but I've written a lot of paragraphs in science

Explanation:

Remember:

  • A paragraph should be in CER formation: Claim (hypothesis/discovery), Evidence, and Reasoning
  • Your reasoning should never start with I believe or I think, your reasoning and evidence are 100% fact based
  • The actual hypothesis should only be 1-2 sentences and be straight to the point

Include:

  • for your hypothesis, keywords relating to your subject, I'm assuming  for your topic you should include things like marking, telescope, stars, etc.
  • why you believe your hypothesis is true
  • so use prior knowledge to justify your hypothesis
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A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy
11Alexandr11 [23.1K]
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
we know that 
         v²=u²+2aS
        2500=2×10×S
        2500÷20 = S
        S= 125 cm
The ramp is 125 cm
8 0
3 years ago
Lakes
Nataliya [291]
I think is between A&B
I think I would answer B
6 0
3 years ago
The length of a football field is closest to<br> (1) 1000 cm (3) 1000 km<br> (2) 1000 dm (4) 1000 mm
Sergeu [11.5K]

For this case, the first thing you should know is that the length of a football field is around 100 meters.

We must then look for a measure close to this value.

We have the following unit conversion:

1 meter = 10 decimeters

Applying the conversion we have:

(1000 dm)*(\frac{1}{10} \frac{m}{dm}) = 100 m

Therefore, the measure closest to a soccer field is:

1000 dm

Answer:

The length of a football field is closest to:

(2) 1000 dm

5 0
3 years ago
Read 2 more answers
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
2 years ago
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