=0.5 x40
<span>= (calculate it) </span>
<span>2. F= -.45 x 41 </span>
<span>= </span>
<span>3. F=20 x 12 </span>
<span>= </span>
<span>4. F=50 x 30 </span>
<span>= </span>
<span>I believe you can go on from here, remember its easy </span>
<span>for the last 2 just rearange the formual </span>
<span>for 9 it will be a=F/m and 10. m=f/a </span>
K = ∆F / ∆s = 4N / 0.08m = 50N/m
Answer:
1.74 m/s
Explanation:
From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.
Therefore, using the formula below, we can calculate the speed of the can, V(can);
===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).
Since the question says the collision was elastic, we use the formula below
Slotting in the given values into the equation (1) above, we have;
1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).
Therefore, final velocity of the can= 2M1V1/M1+M2.
==> 2×2.7×1.1/ 2.7 + 0.72.
The velocity of the can after collision = 1.74 m/s
Evaporation is more contrasted with the rainfall which is less in the correlation of tropical locale. In tropical locale, protection is very little in light of mists. Alongside temperature, it is the main consideration in adding to changes in the thickness of seawater and consequently sea flow. Saltiness is the way to understanding the worldwide water cycle. 97% of the Earth's free water dwells in the seas. The water cycle is commanded by precipitation and evaporation.
To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this
Here,
k = Coulomb's constant
q = Charge
r = Distance to the center point between the charge
From each object the potential will be
Replacing the values we have that
Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,
Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say
Here
m = mass
v = Velocity
q = Charge
V = Voltage
Rearranging to find the velocity
Replacing,
Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is 
