C.) Electron shells or cloud! :)
Answer: 0.306
Explanation:
from the question we are given the following
mass of sled (m) = 50 kg
force (f) = 1.75 x 10^2 N = 175 N
distance (s) = 6 m
net work done on the sled = 1.50 x 10 ^2 N = 150 N
acceleration due to gravity (g) = 9.8 m/s^2
coefficient of friction = μ
lets first calculate the frictional force (ff)
ff = μ x m x g = μ x 50 x 9.8 = 490 μ
work done on the slide by the applied force (W1)= f x s = 175 x 6 = 1050 j
work done on the slide by frictional force (W2) = ff x s = 490 μ x 6 = 2940μ j
now the net work done is the work done by the frictional force subtracted from the work done by the applied force
net work done = W1 - W2
150 = 1050 - 2940μ
2940μ = 1050 - 150
μ = 900 / 2940
μ = 0.306
The Specific Heat Capacity of silver is 230 J/kgK, melting point is 961.8 C so the difference is 941.8K. Now we simply do q=230J/kgK*16.5kg*941.8K and that is 3 574 131 J
The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.
Your answer is: B) Increasing the number of coils around the core
Have an amazing day and stay hopeful!
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1
