Answer:
k1 + k2
Explanation:
Spring 1 has spring constant k1
Spring 2 has spring constant k2
After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.
x1 = x2
Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are
k1 = F1/x -> F1 =k1*x
k2 = F2/x -> F2 =k2*x
While F = F1 + F2
Substitute equation of F1 and F2 into the equation of sum of forces
F = F1 + F2
F = k1*x + k2*x
= x(k1 + k2)
Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)
Considering the general equation of spring forces (Hooke's Law) F = kx,
The effective spring constant for the system is k1 + k2
Answer:
Part a) When collision is perfectly inelastic
Part b) When collision is perfectly elastic
Explanation:
Part a)
As we know that collision is perfectly inelastic
so here we will have
so we have
now we know that in order to complete the circle we will have
now we have
Part b)
Now we know that collision is perfectly elastic
so we will have
now we have
Answer:
Explanation:
2 )
power of an electric device = V² / R where V is volts and R is resistance
putting given data
power = 9²/ 5
= 16.2 J/s
energy produced in 7 minutes
= 16.2 x 7 x 60
= 6804J .
3 ) Power of an electrical device
= V² / R
= V X I where I is current
= 4.5 x .5
= 2.25 W or J/s
4 )
energy used in 3 minutes with power of 2.25 W
= 2.25 x 3 x 60
= 405 J .
7 )
power of a electrical device
= V x I
IR x I where R is resistance .
= I²R
putting given data
power = .005² x 50
= 1.25 x 10⁻³ W .
8 )
Energy used up by a 60 W bulb in 2.5 hours
= 60 x 2.5 x 60 x 60
= 5.4 x 10⁵ J .
Answer:
Use the formula
g = GM/R^2
where
g = acceleration of gravity on moon
G = universal gravitational constant
= 6.67 x 10^-11 m^3/kg-s^2
M = mass of the moon
R = radius of the moon
solving for R, we get
R = 1.74 x 10^6 kg
