Answers
Answer:
a) Revolutions per minute = 2.33
b) Centripetal acceleration = 11649.44 m/s²
Explanation:
a) Angular velocity is the ratio of linear velocity and radius.
Here linear velocity = 72 m/s
Radius, r = 0.89 x 0. 5 = 0.445 m
Angular velocity
Frequency
Revolutions per minute = 2.33
b) Centripetal acceleration
Here linear velocity = 72 m/s
Radius, r = 0.445 m
Substituting
Centripetal acceleration = 11649.44m/s²
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Answers
Answer:
The acceleration expressed in the new units is
Explanation:
To convert from to
it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:
Then by means of a rule of three it is get:
Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:
So the acceleration expressed in the new units is .
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D
Answers
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:
Answers
Final answer:
To find the time rate of change of electric flux between the plates of the capacitor, use the formula \(\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}\). The displacement current between the plates can be found using the formula \(I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}\).
Explanation:
To find the time rate of change of electric flux between the plates of the capacitor, we can use the formula: \(\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}\), where \(\frac{d\phi_E}{dt}\) is the time rate of change of electric flux, \(I\) is the current, and \(A_\text{plate}\) is the area of one plate. In this case, the area of each plate is \((0.06 \,\text{m})^2\) and the current is 0.134 A. Thus, the time rate of change of electric flux is \(\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}\) V·m/s.
The displacement current between the plates of a capacitor can be found using the formula: \(I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}\), where \(I_d\) is the displacement current, \(\varepsilon_0\) is the vacuum permittivity, \(\kappa\) is the dielectric constant, and \(\frac{d\phi_E}{dt}\) is the time rate of change of electric flux. In this case, \(\varepsilon_0\) is a constant, \(\kappa\) depends on the material between the plates (not provided), and we found \(\frac{d\phi_E}{dt}\) to be \(\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}\) V·m/s. So the displacement current can be calculated once these values are known.
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Final answer:
The time rate of change of electric flux can be found using I/ε₀A. The displacement current can be found using ε₀A(dE/dt)
Explanation:
The time rate of change of electric flux between the plates of the capacitor can be found using the formula:
dΦ/dt = I/ε₀A
Where dΦ/dt is the time rate of change of electric flux, I is the current, ε₀ is the permittivity of free space, and A is the area of one of the plates.
We are given that the current is 0.134 A and the area of each plate is (0.060 m)² = 0.0036 m². Plugging these values into the equation, we get: the time rate of change of electric flux between the plates is 37.22 V·m/s.
Similarly, the displacement current between the plates can be found using the formula:
Id = ε₀A(dE/dt)
Where Id is the displacement current, ε₀ is the permittivity of free space, A is the area of one of the plates, and dE/dt is the time rate of change of electric field intensity between the plates.
We are given that ε₀ is 8.854 × 10⁻¹² F/m and dΦ E/ dt is 37.22 V·m/s. Plugging these values into the equation, we get:
Id = (8.854 × 10⁻¹² F/m)(37.22 V·m/s) = 3.29 × 10⁻¹⁰ A
Therefore, the displacement current between the plates is 3.29 × 10⁻¹⁰ A.
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Answers
The correct option is option (1)
The faster movement of air on the upper surface of the paper creates lower pressure above the paper.
Pressure difference:
The movement of air is always from a region of higher pressure to a region of lower pressure.
As we blow air above the paper strip a low pressure is created above the strip due to the fast movement or high speed of the air. And the pressure below the strip is higher in comparison to the pressure above since the air below is not moving.
So, due to the pressure difference, a force is generated on the paper strip by the air from the lower surface to the upper surface.
Learn more about pressure difference:
This is happened because "the air" above "moves faster" and "the pressure" is "lower".
Option: 1
Explanation:
Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.
A. Energy
B. The rope
C. The athlete
D. Air
Answers
Answer:
It would be B. The Rope
Explanation:
I say this because the rope is transferring energy from one location to another. Now, I could be totally wrong on this but I think this is right lol.
Answer:
The answer is B the rope.
Hope this helps <3 ;)
Answers
Answer:
(a)
(b) Initial velocity of the projectile is 22.54 m/s
(c) Straight line perpendicular to the plane of the car's motion
(d) Parabolic
(e) The initial velocity is 23.04 m/s
Solution:
As per the question:
Velocity of the cart, v = 0.500 m/s
Distance moved by the cart, d = 2.30 m
Now,
(a) The projectile must be fired at an angle of so that it mounts on the top of the cart moving with constant velocity.
(b) Now, for initial velocity, u':
Time of flight is given by;
(1)
where
T = Flight time
D = Distance covered
(b) The component of velocity w.r.t an observer:
Horizontal component,
Vertical component,
Also, the vertical component of velocity at maximum height is zero,
Therefore,
Total flight time, (2)
Now, from eqn (1) and (2):
(c) The shape of the projectile w.r.t an observer will be a straight line perpendicular to the plane of cart's motion.
(d) The shape of the path of the projectile seen by the physics student outside the reference frame of the cart is parabolic
(e) The initial velocity is given by:
u = u' + v = 22.54 + 0.5 = 23.04 m/s