Answer:
option C
Explanation:
The correct answer is option C.
The normal force is the force exerted by the biker on the inner vertical surface of the circular track.
When the biker move in the circular track centripetal force is acting on the biker which is being balanced by the normal force.
To overcome the gravitation force on the biker the velocity of the biker should be high such that centripetal acceleration of the biker can overcome the gravity force acting on the biker.
Answer:
Option C is the correct answer.
Explanation:
We have ideal gas equation, PV = nRT
Volume of tank, V = 1 m³
Mass of water = 10 kg
Mass of 1 mole of water = 18 g = 0.018 kg
Number of moles,
![n=\frac{10}{0.018}=555.56moles](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B10%7D%7B0.018%7D%3D555.56moles)
Temperature, T = 160 °C = 160 + 273 = 433 K
Ideal gas constant, R = 8.314 JK⁻¹mol⁻¹
Substituting
P x 1 = 555 x 8.314 x 433
P = 2000000 Pa = 2000 kPa
The pressure in the tank is 2000 kPa
Option C is the correct answer.
Answer:
The truck must travel by 95.16 km/h to stay beneath the airplane
Explanation:
Lets explain how to solve the problem
How fast must a truck travel to stay beneath an airplane that is moving
105 km/h at an angle of 25° to the ground
The truck moves a long the horizontal ground
The airplane moves 105 km/h at an angle 25° to the ground
The truck must moves with velocity equal the horizontal component
of the velocity of the airplane to stay beneath the airplane
The angle between the line represented the velocity of the airplane
and the ground is 25°
→ The horizontal component of the velocity = v cos(25)
→ The velocity of the airplane is 105 km/h
→ The horizontal component = 105 cos(25) = 95.16 km/h
The velocity of the truck = the horizontal component of the airplane
<em>The truck must travel by 95.16 km/h to stay beneath the airplane</em>
Answer: true
Explanation: If the mass of an object is increased so is the gravitational pull
Answer:
The electric force between them if the pieces of grain are 2 cm apart is
.
Explanation:
Given:
Charge on one grain, ![Q_{1}=6\times 10^{-10}\textrm{ C}](https://tex.z-dn.net/?f=Q_%7B1%7D%3D6%5Ctimes%2010%5E%7B-10%7D%5Ctextrm%7B%20C%7D)
Charge on another grain, ![Q_{2}=2.3\times 10^{-15}\textrm{ C}](https://tex.z-dn.net/?f=Q_%7B2%7D%3D2.3%5Ctimes%2010%5E%7B-15%7D%5Ctextrm%7B%20C%7D)
Separation between them, ![d=2\textrm{ cm}=0.02\textrm{ m}](https://tex.z-dn.net/?f=d%3D2%5Ctextrm%7B%20cm%7D%3D0.02%5Ctextrm%7B%20m%7D)
Electric force acting between two charges
separated by a distance
is given as:
![F_{elec}=\frac{kQ_{1}Q_{2}}{d^2}](https://tex.z-dn.net/?f=F_%7Belec%7D%3D%5Cfrac%7BkQ_%7B1%7DQ_%7B2%7D%7D%7Bd%5E2%7D)
Where,
is Coulomb's constant equal to
.
Now, plug in all the values and solve for
.
![F{elec}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}](https://tex.z-dn.net/?f=F%7Belec%7D%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%206.0%5Ctimes%2010%5E%7B-10%7D%5Ctimes%202.3%5Ctimes%2010%5E%7B-15%7D%7D%7B%280.02%29%5E2%7D%5C%5C%5C%5CF_%7Belec%7D%3D3.15%5Ctimes%2010%5E%7B-11%7D%5Ctextrm%7B%20N%7D)
Therefore, the electric force between them if the pieces of grain are 2 cm apart is
.