Answers
Answer:
4 m/s^2
Explanation:
(0.5m/s * 4)+2 m/s
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3.what does this stand for ??
Thank you!!
Answers
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kN
b. Left support = 36 kN
Right support = 29 kN
c. Left support force will decrease
Right support force will increase.
d. Right support = 43 kN
Explanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
Answers
Answer:
D = 58 cm
Explanation:
Given that,
Focal length of the objective lens,
Focal length of the eye piece,
We need to find how many cm apart should the lenses be placed. Let d be the distance between lenses. It is equal to the sum of focal lengths of objective lens and eye-piece
D = 48 cm + 10 cm
= 58 cm
Hence, the object is placed at a distance of 58 cm.
Final answer:
In an astronomical telescope, the lenses should be placed at a distance equal to the sum of their focal lengths. In this case, this distance would be 58 cm.
Explanation:
In an astronomical telescope, the distance between the objective lens and the eyepiece should be equal to the sum of their focal lengths for the telescope to produce clear and sharp images. Here, the focal length of the objective lens is 48 cm and the focal length of the eyepiece is 10 cm.
Therefore, calculating: Objective lens focal length + Eyepiece focal length = 48 cm (objective) + 10 cm (eyepiece) = 58 cm
This means that the objective lens and the eyepiece should be approximately 58 centimeters apart.
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Answers
Answer:
37.5 J
Explanation:
With work done equation: W=Fs
W=75*0.50=37.5 J
or use mgh=(75)(0.5) which is the same
Answers
Answer:
The energy stored is
Explanation:
From the question we are told that
The capacitance is
The resistance is R = 3.00-Ω
The emf is
The power is P = 300 W
Generally the total emf is mathematically represented as
Here is the emf across that capacitor which is mathematically represented as
and is the emf across the resistor which is mathematically represented as
So
=>
Generally the energy stored in a capacitor is mathematically represented as
=>
=>
=>
Final answer:
The energy stored in the capacitor is 0 J.
Explanation:
When a 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V
At the instant when the resistor is dissipating electrical energy at a rate of 300 W, we can calculate the current flowing through the circuit using Ohm's law: I = V/R = 70.0 V / 3.00 Ω = 23.33 A.
The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.
Since the capacitor is initially uncharged, the voltage across it is also zero. So the energy stored in the capacitor is 0.5 * 8.00 x 10^-6 F * (0 V)^2 = 0 J.
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Answers
Answer:
Explanation:
initial velocity, u = 25 m/s
distance, s = 55 m
coefficient of static friction = 0.6
coefficient of kinetic friction = 0.3
Let the acceleration is a.
Use third equation of motion
v² = u² + 2as
0 = 25 x 25 - 2 x a x 55
a = 5.68
a = μg
μ = 5.68 / 9.8 = 0.58
so, the coefficient of friction is less then the coefficient of static friction so the antiques are safe.
Answers
Answer:
See the explaination for the details.
Explanation:
Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
Please kindly check attachment for the step by step explaination of the answer.