Answer:
the answer is 3.3 %
Explanation:
Answer:
k = 1/18
Explanation:
Data:
a = -k
to = 0s Vo = 12m/s
t = 6s the particle chage it's moviment, so v = 0 m/s
We know that acceleration is the derivative of velocity related to time:
rearranging...
Then, we must integrate both sides:
V = 0 because the exercise says that the car change it's direction:
k = 1/6
In order to find X - Xo we must integer v*dT = dX
so...
integrating...
X - Xo = 54m
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D
The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.
The gravitational force between mass A and mass D is therefore:
F = G * m_A * m_D / r²
= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²
= 1.15 × 10⁻⁷ N
The direction of the gravitational force is towards mass D.
Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.
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Answer:
THE ANSER IS B
Explanation:
Given Information:
diameter = d = 15 mm
Length = L = 20 mm
Axial load = P = 300 N
Eₚ = 2.70x10⁹ Pa
vₚ = 0.4
Required Information:
Change in length = ?
Change in diameter = ?
Answer:
Change in length = 0.01257 mm
Change in diameter = -0.003772 mm
Explanation:
Stress is given by
σ = P/A
Where P is axial load and A is the area of the cross-section
A = 0.25πd²
A = 0.25π(0.015)²
A = 0.000176 m²
σ = 300/0.000176
σ = 1697792.8 Pa
The longitudinal stress is given by
εlong = σ/Eₚ
εlong = 1697792.8/2.70x10⁹
εlong = 0.0006288 mm/mm
The change in length can be found by using
δ = εlong*L
δ = 0.0006288*20
δ = 0.01257 mm
The lateral stress is given by
εlat = -vₚ*εlong
εlat = -0.4*0.0006288
εlat = -0.0002515 mm/mm
The change in diameter can be found by using
Δd = εlat*d
Δd = -0.0002515*15
Δd = -0.003772 mm
Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm
Answer:
-0.00152 V
Explanation:
Parameters given:
Diameter of the loop = 11 cm = 0.11m
Rate of change of magnetic field, dB/dt = 0.16 T/s
Radius of the loop = 0.055m
The area of the loop will be:
A = pi * r²
A = 3.142 * 0.055²
A = 0.0095 m²
The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:
EMF = - N * A * dB/dt
In this case, there's only one loop, so N = 1.
Therefore:
EMF = -1 * 0.0095 * 0.16
EMF = -0.00152 V
The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.
Answer:
The induced emf is 0.00152 V
Explanation:
Given data:
d = 11 cm = 0.11 m
The area is:
The induced emf is:
The negative indicates the direction of E.
To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam, we need to consider both the change in temperature and the phase change from steam to liquid. The specific heat of water is used to calculate the heat required to lower the temperature, while the heat of vaporization is used to calculate the heat required to condense the steam. Adding these two heat values together gives us the total amount of heat that must be removed from the steam, which is approximately 3.61164 kcal.
When steam at 100°C condenses and its temperature is lowered to 46°C, heat must be removed from the steam. To calculate the amount of heat, we can use the specific heat of steam and the latent heat of vaporization. First, we calculate the heat required to lower the temperature of the steam from 100°C to 46°C using the specific heat of water. We then calculate the heat required to condense the steam using the latent heat of vaporization. Finally, we add these two heat values together to obtain the total amount of heat that must be removed from the steam.
Given:
Calculations:
Calculation:
Therefore, the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C is approximately 3.61164 kcal.
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To condense and cool 6.1 g of 100°C steam to 46°C, 3.2879 kcal must be removed for condensation, and 0.3304 kcal for cooling, for a total of 3.6183 kcal.
Calculating the Quantity of Heat for Condensation and Cooling
To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C, we need to consider two processes: condensation and cooling. For condensation, we use the heat of vaporization, and for cooling, we use the specific heat of water.
Total heat removed is the sum of the heat from both steps: 3.2879 kcal + 0.3304 kcal = 3.6183 kcal.
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B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s
Answer:
A. A tractor trailer rig moving at 2 m/s
Explanation:
Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.
In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.
The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a tractor trailer rig, the greater will be its tendency to continuously remain at rest.
Hence, the object that has more inertia is a tractor trailer rig moving at 2 m/s because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia.