Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
The density of an object can be calculated using the formula Density = Mass/Volume.
Experimental Density:
Density = 153.8g / 20.00 cm^3
Density = 7.69g/cm^3
Percent error equation:
% Error = | Theoretical Value - Experimental Value|/Theoretical Value * 100
% Error = | 7.87g/cm^3 - 7.69g/cm^3|/7.87g/cm^3 * 100
% Error = 2.29%
Therefore a is the correct answer.
By Boyle's law the volume of the sample decreases, provided temperature is constant.
Explanation:
Load=800N
Effort=200N
1. Mechanical Advantage = LOAD/EFFORT
= 800N/200N
= 4
2 Velocity Ratio = no. Of pulleys =5
3. Efficiency = Mechanical advantage / velocity ratio × 100%
= (4/5)×100%
=80%
4. output work= load×load distance
= 800N × 5m
= 4 × 1000J
5. Efficiency = (output work/input work) ×100%
Or, 80% = (4000J/input work) ×100%
Or, 80%/100% = 4000J/inputwork
Or, 4/5 = 4000J/inputwork
Or, input work =4000J × 5/4
Input work = 5×1000J
I hope it helped! ;-)
Answer:
temperature change is 262.06°K
Explanation:
given data
mass = 0.07 kg
velocity = 258 m/s
to find out
what is its temperature change
solution
we know here
heat change Q is is equal to kinetic energy that is
KE = 0.5 × m× v² ...........1
here m is mass and v is velocity
KE = 0.5 × 0.07 × 258²
KE = 2329.74 J
and we know
Q = mC∆t .................2
here m is mass and ∆t is change in temperature and C is 127J/kg-K
so put here all value
2329.74 = 0.07 × 127 × ∆t
∆t = 262.06
so temperature change is 262.06°K
