Answer:
Explanation:
The amplitude of the oscillation under SHM will be .5 m and the equation of
SHM can be written as follows
x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.
x = .5 cosωt
given , when t = .2 s , x = .35 m
.35 = .5 cos ωt
ωt = .79
ω = .79 / .20
= 3.95 rad /s
period of oscillation
T = 2π / ω
= 2 x 3.14 / 3.95
= 1.6 s
b )
ω = 
ω² = k / m
k = ω² x m
= 3.95² x .6
= 9.36 N/s
c )
v = ω
At t = .2 , x = .35
v = 3.95 
= 3.95 x .357
= 1.41 m/ s
d )
Acceleration at x
a = ω² x
= 3.95 x .35
= 1.3825 m s⁻²
Answer:
The speed is 
.
(a) is correct option.
Explanation:
Given that,
Potential difference 
Speed 
If it were accelerated instead
Potential difference 
We need to calculate the speed
Using formula of initial work done on proton
We know that,
Put the value into the formula
....(I)
If it were accelerated instead through a potential difference of 
, then it would gain a speed will be given as :
Using an above formula,
Put the value of 
Hence, The speed is .
Answer:
Required heat Q = 11,978 KJ
Explanation:
Given:
Mass = 5.3 kg
Latent heat of vaporization of water = 2,260 KJ / KG
Find:
Required heat Q
Computation:
Required heat Q = Mass x Latent heat of vaporization of water
Required heat Q = 5.3 x 2260
Required heat Q = 11,978 KJ
Required heat Q = 12,000 KJ (Approx.)
Because there is no record of all things. As we have partially unknow information, it can never be held as a fact.
Answer:
The time it took the bobsled to come to rest is 10 s.
Explanation:
Given;
initial velocity of the bobsled, u = 50 m/s
deceleration of the bobsled, a = - 5 m/s²
distance traveled, s = 250 m
Apply the following kinematic equation to determine the time of motion of the bobsled;
s = ut + ¹/₂at²
250 = 50t + ¹/₂(-5)t²
250 = 50t - ⁵/₂t²
500 = 100t - 5t²
100 = 20t -t²
t² - 20t + 100 = 0
t² -10t - 10t + 100 = 0
t (t - 10) - 10(t - 10) = 0
(t - 10)(t - 10) = 0
t = 10 s
Therefore, the time it took the bobsled to come to rest is 10 s.
