An RV is traveling 60 km/h along a highway. A boy sitting near the driver of the RV throws a ball to another boy at the back end
1 answer:
Answer:
Speed of the ball relative to the boys: 25 km/h
Speed of the ball relative to a stationary observer: 35 km/h
Explanation:
The RV is travelling at a velocity of
Here we have taken the direction of motion of the RV as positive direction.
The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is
with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is
v = 25 km/h
Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:
v = +60 km/h + (-25 km/h) = +35 km/h
So, he/she measures a speed of 35 km/h.
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a) Electric field inside the paint layer: zero
b) Electric field just outside the paint layer:
c) Electric field 8.00 cm outside the paint layer:
Explanation:
a)
We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:
where
E is the electric field
dS is the element of surface
q is the charge within the gaussian surface
is the vacuum permittivity
Here we want to find the electric field just inside the paint layer, so we take a sphere of radius as Gaussian surface, where
R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)
By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore
So we have
which means that the electric field inside the paint layer is zero.
b)
Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius
The area of the surface is
And since the electric field is perpendicular to the surface at any point, Gauss Law becomes
The charge included within the sphere in this case is the charge on the paint layer, therefore
So, the electric field is:
where the negative sign means the direction of the field is inward, since the charge is negative.
c)
Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.
Therefore, we have to take a Gaussian sphere of radius:
Gauss theorem this time becomes
And the charge included within the sphere is again the charge on the paint layer,
Therefore, the electric field is
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The final temperature of the system is 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry
The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x
x (T - 10)
⇒ 14,850 - 330T = 183.21T - 1832
⇒ - 513.21 T = - 16682
or T = 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry
The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x
⇒ 14,850 - 330T = 183.21T - 1832
⇒ - 513.21 T = - 16682
or T = 32.5°
Answer:
T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
∑ F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
cos 41 = T₁ₓ / T1
sin 41 = T₁y / T1
T₁ₓ = T₁ cos 41
T₁y= T₁ sin 41
for cable gold
cos 63 = T₂ / T₂
sin 63 = / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
T₃ - W = 0
T₃ = W
T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
- T₁ₓ + T₂ₓ = 0
T₁ₓ = T₂ₓ
T1 cos 41 = T2 cos 63
T1 = T2 cos 63 / cos 41 (1)
y Axis
+ T_{2y} - T3 = 0
T₁ sin 41 + T₂ sin 63 = T₃ (2)
to solve the system we substitute equation 1 in 2
T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
T₂ (cos 63 tan 41 + sin 63) = W
T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
T₂ = 200 / (cos 63 tan 41 + sin 63)
T₂ = 200 / 1,2856
T₂ = 155.6 N
we substitute in 1
T₁ = T₂ cos 63 / cos 41
T₁ = 155.6 cos63 / cos 41
T₁ = 93.6 N
therefore the tension in each cable is
T₁ = 93.6 N
T₂ = 155.6 N
T₃ = 200 N