Answer:
A:1.94
Explanation:
cause that the only one on there
Answer:
Speed of the car 1 =
Speed of the car 2 =
Explanation:
Given:
Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)
mathematically,
M₁ = 2M₂
Kinetic Energy of the car 1 = Half the kinetic energy of the car 2
KE₁ = 0.5 KE₂
Now, the kinetic energy for a body is given as
where,
m = mass of the body
v = velocity of the body
thus,
or
or
or
or
or
.................(1)
also,
or
or
or
or
or
or
or
or
and, from equation (1)
Hence,
Speed of car 1 =
Speed of car 2 =
Answer:
Direction of ship: 9.45° West of North
Ship's relative speed: 7.87m/s
Explanation:
A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0
Vx=0
Therefore, -VsSin∅+VcCos∅40°
Sin∅ = Vc/Vs × Cos 40°
Sin∅ = 1.5/7 ×Cos40°
Sin∅= 0.164
∅= Sin-¹ (0.164)
∅= 9.45° W of N
B. Ship's relative speed:
Vy= VsCos∅ + Vcsin40°
= 7Cos9.45° + 1.5sin40°
= 7×0.986 + 1.5×0.642
= 7.865
= 7.87m/s
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
-----------------------------------------------------------------
20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.
The linear mass density is given as,
The expression for the wavelength of the standing wave for the second overtone is
Replacing we have
The frequency of the sound wave is
Now the velocity of the wave would be
The expression that relates the velocity of the wave, tension on the string and linear mass density is
The tension in the string is 547N
PART B) The relation between the fundamental frequency and the harmonic frequency is
Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore
Then,
Rearranging to find the fundamental frequency