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sergij07 [2.7K]
2 years ago
12

SORRY! COULDN'T EDIT MORE INFO ON MY LAST QUESTION! RE ASKING

Physics
1 answer:
tiny-mole [99]2 years ago
8 0

The position isn't changing ! The object is not moving. It's just sitting there. No matter what 't' you're interested in, the position is exactly the same as it has been since t=0 .


That's (4 divisions) x (2.8 m/div). I'm sure you can calculate the number of meters.

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Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T
Westkost [7]

Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

We know that the kinetic energy of any particle is given by :

KE=\dfrac{1}{2}mv^2

Where,

m is the mass of an object.

v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

Hence, the conclusion of the question is if the mass of a particle is increases then its kinetic energy also increase.

3 0
3 years ago
Read 2 more answers
1.Convert 340 cm into m *(answer=0.34m)
Nataly [62]

Answer:

<em>1</em><em>.</em><em>for </em><em>the </em><em>first </em><em>one </em><em>100c</em><em>e</em><em>n</em><em>t</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em> </em><em>make </em><em>1</em><em> </em><em>meter </em><em>therefore</em>

<em>100c</em><em>m</em><em>-</em><em>1</em><em>m</em>

<em>3</em><em>4</em><em>0</em><em>c</em><em>m</em><em>-</em><em>x</em>

<em>3</em><em>4</em><em>0</em><em>/</em><em>100</em>

<em>=</em><em>3</em><em>.</em><em>4</em>

<em>the </em><em>answer </em><em>is </em><em>supposed</em><em> to</em><em> be</em><em> </em><em>3</em><em>.</em><em>4</em><em>,</em><em> maybe</em><em> </em><em>there's</em><em> </em><em>a </em><em>mistake</em><em> </em><em>with </em><em>the </em><em>question</em><em> </em><em>or </em><em>the </em><em>answer</em>

<em>2</em><em>.</em><em>t</em><em>h</em><em>e</em><em> </em><em>weight</em><em> </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given </em><em>by </em><em>the </em><em>formula</em>

<em>mass×</em><em>g</em><em>r</em><em>a</em><em>v</em><em>i</em><em>t</em><em>y</em><em>,</em><em>in </em><em>this </em><em>case </em><em>the </em><em>mass </em><em>is </em><em>7</em><em>5</em><em>k</em><em>g</em><em> </em><em>and </em><em>the </em><em>gravity </em><em>is </em><em>9</em><em>.</em><em>8</em>

<em>weight</em><em>=</em><em>7</em><em>5</em><em>×</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>3</em><em>5</em><em>N</em>

<em>3</em><em>.</em><em>f</em><em>o</em><em>r</em><em> </em><em>this </em><em>one </em><em>the </em><em>mass </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given</em><em> by</em><em> the</em><em> formula</em>

<em>mass=</em><em>weight/</em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>0</em><em>/</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>.</em><em>8</em><em>k</em><em>g</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

4 0
3 years ago
Match each chemical name with the correct chemical formula.
pochemuha
Is there any chemical names listed ?
4 0
3 years ago
A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0
3 years ago
The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G
Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

7 0
3 years ago
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