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MaRussiya [10]
3 years ago
7

Suppose that a constant force is applied to an object. Newton's Second Law of Motion states that the acceleration of the object

varies inversely with its mass. A certain force acting upon an object with mass 2kg produces an acceleration of 10ms^2 . If the same force acts upon another object whose mass is 5kg , what would this object's acceleration be?
Physics
1 answer:
omeli [17]3 years ago
4 0
<span>(9 kg)(5 m/s^2) = M(3 m/s^2) 
</span><span>that the acceleration of the object varies inversely with its mass.</span>
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A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga
pashok25 [27]
A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
p_i V_i = nRT_i
where
p_i=1.0 atm=1.01 \cdot 10^5 Pa is the initial pressure of the gas
V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
5 0
3 years ago
Neutral hydrogen can be modeled as a positive point charge +1.6×10^−19C surrounded by a distribution of negative charge with vol
Archy [21]

Answer:

a) 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

4 0
3 years ago
9. If the frequency of a certain light is 3.8 x 1024 Hz, what is the energy of this light?
german

Answer:

E=hf

Were, h = Planck constant 6.67*10^11

E=3.8*10^24 * 6.67*10^11= 2.508*q0^36j

4 0
3 years ago
Michael's house is 5.0 km away from his school. How long would it take him to go ti school, riding a bus, if its velocity is 25
Anvisha [2.4K]

Answer:

12 mins

Explanation:

The distance covered is 5 km, divide this by 25 to get the fraction of an hour it takes. Doing this you get .2, times this by 60 min (1 hour) to get how many mins it takes

8 0
3 years ago
What does it mean to say the mass is conserved during a physical change ?
Olin [163]
Mater doesn't just appear or disappeared. Chemical elements are still there just the connections and how it combines changes.

So what goes into your chemical eqation must still exist after the change.
4 0
4 years ago
Read 2 more answers
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