relation between potential difference and electric field is given as
so here we know that
d = 3 cm
So now when plates are separated to 4 cm distance carefully
the potential difference between them will change but the electric field between them will remain constant
So at distance of 4 cm also the electric field will be E = 1000 N/C
Answer:
t = 2.77 s
Explanation:
The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:
Equation of movement of the ball in the X axis
X = v₀x*t Equation (1)
Equation of movement of the ball in the Y axis
Y = y₀+ v₀y*t -½ g*t² Equation (2)
Where
X : horizontal position in meters (m)
Y : vertical position in meters (m)
y₀ : initial vertical position in meters (m)
v₀x : X-initial speed in m/s
v₀y : Y-initial speed in m/s
g: acceleration due to gravity in m/s²
t : time to position (X,Y)
Data
y₀ = 37.5 m
v₀y = 0
g = 9.8 m/s²
Problem development
The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0
We apply the Equation (2):
Y = y₀+ (v₀y)*t - (½) g*t²
0 = 37.5 +(0)*t- (1/2)*(g)*t²
0 = 37.5 - (1/2)*(9.8)*t²
(1/2)*(9.8)*t² = 37.5
t² = (2)(37.5)/(9.8)
t = 2.77 s
Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v= \
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
Answer:
A) 10 m/s
Explanation:
We know that according to conservation of momentum,
m1v1 + m2v2 = m1u1 + m2u2 ..............(equation 1)
where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.
From the given data
If truck and car are two bodies
truck : m1 = 2000 Kg v1 = 5 m/s u1 = 0
car : m2 = 1000 kg v2 = 0 u2 = ?
final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.
substituting the values in equation 1, we get
(2000 x 5) + 0 = 0 + (1000 x u2)
u2 = x 5
= 10 m/s
Hence after collision, car moves at a velocity of 10 m/s