Complete Question
A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.
Answer:
The value for static friction is 
The value for static friction is 
Explanation:
From the question we are told that
The mass of the clock is 
The first horizontal force is 
The second horizontal force is 
Generally the static frictional force is equal to the first horizontal force
So
=> 
=>
Generally the kinetic frictional force is equal to the second horizontal force
So
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
Answer:
150.8 J
Explanation:
The heat released by the copper wire is given by:
where:
m = 10.0 g is the mass of the wire
Cs = 0.377 j/(g.C) is the specific heat capacity of copper
is the change in temperature of the wire
Substituting into the equation, we find
And the sign is negative because the heat is released by the wire.
Answer:
The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m
Explanation:
Data
F = 2 N
d = ?
q = 1.6 x 10 ⁻¹⁹ C
k = 8.987 Nm²/C²
Formula
Solve for r
Substitution
Simplification
r = 
r = 
Result
r = 1.07 x 10⁻¹⁴ m
