Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
Answer:
1. **Position vs. Time Plot and Acceleration**:
To create a position vs. time plot, we can use the data provided. I'll first list the data in a table format:
| t (s) | x (m) |
|-------|-------|
| 0 | 0 |
| 0.05 | 1 |
| 1 | -12 |
| 3 | -40 |
| 5 | -105 |
| 8 | -175 |
| 10 | 15 |
| 22 | -2060 |
| 33 | ??? |
However, there seems to be a discrepancy in the data at t = 10. The position is given as both 15 and -410 -900. Please clarify this point, and I'll calculate the acceleration once we have the correct data.
2. **Estimate Number of Breaths in One Week**:
To estimate the number of breaths a person takes in one week, we can make some assumptions:
- An average person takes about 12-20 breaths per minute at rest.
- Let's assume 15 breaths per minute on average.
- In an hour, that's 15 breaths/minute * 60 minutes/hour = 900 breaths.
- In a day, it's 900 breaths/hour * 24 hours/day = 21,600 breaths.
- In a week (7 days), it's 21,600 breaths/day * 7 days/week = 151,200 breaths in one week.
3. **Narrative of Motion and Calculations**:
I'll need more information about the graph you mentioned for part 3. Could you describe the graph or provide the relevant data or equations related to it? This will allow me to answer parts a, b, c, and d accurately.
A) 450 meters
B) 495 meters
C) 4.09 meters
D) 498 meters
Data given:
V=45m/s
t=11s
Δx=?
Formula needed:
V=Δx/t
Solution:
Δx=v×t
Δx=45m/s×11s
Δx=495m
According to my solution B) is the most accurate
Answer:22.76 m/s
Explanation:
Given
Train length(L)=75 m
Front of train after travelling 125 m is 18 m/s
Time taken by the front of train to cover 125 m
Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel a distance of 125+75=200 m
Answer:
E₁ / E₂ = M / m
Explanation:
Let the electric field be E₁ and E₂ for ions and electrons respectively .
Force on ions = E₁ e where e is charge on ions .
Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0
v² = u² - 2as
0 = u² - 2 x E₁ e d / M
u² = 2 x E₁ e d / M
Similarly for electrons
u² = 2 x E₂ e d / m
Hence
2 x E₁ e d / M = 2 x E₂ e d / m
E₁ / E₂ = M / m
The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.
The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:
Work = Change in kinetic energy
The change in kinetic energy can be calculated using the formula:
Change in kinetic energy = (1/2)Mv2 - (1/2)mv2
where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:
Ratio = (1/2)M/(1/2)m = M/m
So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.
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( a= ? m/s2 )
Calculate the proton's speed after 1.40 {\rm \mu s} in the field, assuming it starts from rest.
( V= ? m/s )"
To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.
PART A ) For the electrostatic force we have that is equal to
Here
q= Charge
E = Electric Force
PART B) Rearrange the expression F=ma for the acceleration
Here,
a = Acceleration
F = Force
m = Mass
Replacing,
PART C) Acceleration can be described as the speed change in an instant of time,
There is not then
Rearranging to find the velocity,
The magnitude of the electric force felt by the proton is 4.4 x 10^-16 N. The proton's acceleration is 2.64 x 10^11 m/s^2. The proton's speed after 1.40 μs in the field is 3.70 x 10^5 m/s.
The charge of a proton is 1.6 x 10-19 coulombs and the electric field strength is 2750 N/C. Therefore, the magnitude of the electric force felt by the proton is (1.6 x 10-19 C)(2750 N/C) = 4.4 x 10-16 N. The mass of a proton is approximately 1.67 x 10-27 kilograms. Therefore, the proton's acceleration is (4.4 x 10-16 N)/(1.67 x 10-27 kg) = 2.64 x 1011 m/s2. Since the proton starts from rest, its initial velocity (u) is 0. Therefore, the proton's speed after 1.40 μs is v = (2.64 x 1011 m/s2)(1.40 x 10-6 s) = 3.70 x 105 m/s.
#SPJ3