The density of the material would be
25/6 grams per cm^3.
to obtain the result above this is what we do:
density is calculated as: (the mass of the given material or object) / volume of the material
which leads us to 50grams /12cm^3
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
Stress = F / A force per unit area
A = 3.00 cm^2 = 3 E-4 m^2
F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied
F/3 = 2.4E4 N if force not to exceed limit (= f)
f = M a
a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g
Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So

As we find acceleration.Now we need to find time
So

Now for distance
So
