Answer:
a) 45 s , b) vₐ = 90 m / s, v_b = 162 m / s, c) x_b = 3.328 10⁴ m
Explanation:
We can solve this exercise using the kinematic relations
Vehicle A
xₐ = v₀ₐ t + ½ aₐ t²
vehicle B
starts two seconds later
x_b = v_{ob} (t-2) + ½ a_b (t-2) ²
as cars start from rest their initial velocities are zero
at the point where they meet, the position must be the same for both vehicles
xa = 0 + ½ aₐ t²
xb = 0 + ½ a_b (t-2) ²
½ aₐ t² = ½ a_b (t-2) ²
t = (t-2)
t (1 - \sqrt{ \frac{a_a}{a_b} }) = 2
t (1 - ⅔, ) = 2
t = 2 / 0.4444
t = 45 s
b)
the speed of each car
vₐ = voa + aa t
vₐ = 0 + 2 45
vₐ = 90 m / s
v_b = 3.6 45
v_b = 162 m / s
c) xb = 0 + ½ ab (t-2) ²
x_b = ½ 3.6 (45-2) ²
x_b = 3.328 10⁴ m
Answer:
Electric force, F = 20.16 N
Explanation:
It is given that,
Charge on gold nucleus,
An electron is sent at high speed toward a gold nucleus,
Distance between electron and gold nucleus,
We need to find the electric force acting on the electron. The formula for electrostatic force is given by :
F = 20.16 N
The electric force acting on the electron is 20.16 N. Hence, this is the required solution.
Answer:
First option
Explanation:
If the ball is running in a circular motion then its velocity <em>v</em> will be tangency to the circular path.
In this problem the centripetal force that allows the circular movement is the tension <em>T</em> of the rope to which the ball is tied, if the child releases the rope then this tension becomes equal to zero and the circular movement is interrupted.
As the speed <em>v</em> of the ball is always tangential to the circumference at any point of the same, then at the instant in which the rope is released, the ball will follow the same trajectory that it had at that moment, that is, tangential to the circumference.
Observe the attached image.
Therefore the answe is: tangent to the circle
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K