Answer:
f = 6.66 cm
Explanation:
For this exercise we will use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
the expression for magnification is
m = h '/ h = - q / p
with this we have a system of two equations with two unknowns, in the problem they give us the distance to the image q = 10 cm and a magnification of m = -0.5
-0.5 = - q / p
p = - q / 0.5
p = - 10 / 0.5
p = 20 cm
now we can with the other equation look for the focal length
1 / f = 1/20 + 1/10
1 / f = 0.15
f = 6.66 cm
Answer: t = 10 s , s = 2,5 m
Explanation: (1) Speed v = v0 + at and (2) distance s = v0t + ½at².
Now v0 = 0.5 m/s , v = 0 and a = -0.05 m/s²
Solve from equation (1) t = -v0 / a = - 0.5 m/s / -0.05 m/s² = 10 s
Add t = 10 s to equation (2 ) s = 0.5 m/s · 10 s + 0.5· (-0.05 m/s²)· (10 s)²
= 5 m - 2,5 m = 2,5 m
Answer:
Explanation:
a ) When 1 kg water is boiled at constant pressure of 1 atm , its volume increases by following volume
(.824 - .001 )m³
.823 m³
work done by steam = increase in volume x pressure
.823 x 10⁵ J
Heat added
= latent heat of vaporization x mass
= 2260000 J x 1
= 22.6 x 10⁵ J
Increase in internal energy of gas
= heat added - work done by gas
= (22.6 - .823) x 10⁵ J
= 21.777 x 10⁵ J .
Change in Momentum = mv - mu.
u = 0, v = 10 m/s. Note ball accelerated from rest, so initial velocity = 0. u =0
Change in Momentum = mv - mu = 3*10 - 3*0 =30.
Change in Momentum = 30 kgm/s.
Answer with Explanation:
We are given that
Spring constant=k=18 N/m
Amplitude,A=6.5 cm=
1m=100 cm
Number of oscillations=N=19
Time=t=18 s
Frequency=Number of oscillations per second=
Frequency=
1kg=1000g
Mass of the ball=410 g
Maximum speed of the ball=)
Maximum speed of the ball=
Hence, the maximum speed of the ball=0.43 m/s