Answer:
Time taken by the coin to reach the ground is 1.69 s
Given:
Initial speed, v = 11.8 m/s
Height of the building, h = 34.0 m
Solution:
Now, from the third eqn of motion:
Now, time taken by the coin to reach the ground is given by eqn (1):
v' = v + gt
Answer:
The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.
Explanation:
The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.
Also notice that convex mirror always makes virtual images.
Another feature of the convex mirror is that an upright image is always formed by the convex mirror.
An important mirror formula to remember which is applicable for both convex and mirrors
Here:
'u' is an object which gets placed in front of a spherical mirror of focal
length 'f' and image 'u' is formed by the mirror.
Answer:
right side up
Explanation:
To find the critical angle, we need to consider the forces acting on the system. The weight and frictional force must be taken into account. By equating the forces and solving for the critical angle, we can determine at what angle the system just begins to move.
To determine the critical angle for the system shown, we need to consider the forces acting on the objects. The force pulling m1 downwards is its weight, which is equal to its mass multiplied by the acceleration due to gravity. The force preventing m1 from moving is the frictional force, which is equal to the coefficient of friction multiplied by the normal force. The normal force is the force exerted by the surface perpendicular to it, which is equal to the weight of m2 minus the weight of the hanging part of the rope.
At the critical angle, the force of friction is at its maximum value, which is equal to the coefficient of friction multiplied by the normal force. The force pulling m1 downwards is equal to the force of friction. By equating these forces and solving for the critical angle, we can find the answer.
#SPJ2
Answers:
a) 222.22 m/s
b) 800.00 km/h
Explanation:
The speed of a wave is given by the following equation:
Where:
is the speed
is the frequency, which has an inverse relation with the period
is the wavelength
Solving with the given units:
This is the speed of the wave in km/h
Transforming this speed to m/s:
This is the speed of the wave in m/s
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules