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Dmitriy789 [7]
3 years ago
10

Laws of vector addition​

Physics
2 answers:
oee [108]3 years ago
8 0

Answer:

laws ig .

Explanation:

kvv77 [185]3 years ago
8 0
Not quite sure what you’re looking for.

Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.
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andriy [413]
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8 0
3 years ago
Write down the use of hydraulic lift.​
Ksivusya [100]

Answer: The Many Uses for Hydraulic Lifts. Hydraulic lifts are utilized in many extraordinary applications. They may be located in the automotive, shipping, construction, waste removal, mining, and retail industries as they are a powerful approach to elevating and reducing people, goods, and equipment.

8 0
4 years ago
An accelerating voltage of 2.42 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizont
Pavlova-9 [17]

Answer:

Part (a)  The magnitude of the deflection of electron beam on the screen due to the Earth's gravitational field is 5.97*10^{-16}m.

Part (b) The magnitude of the deflection of electron beam on the screen due to the vertical component of the Earth's magnetic field is 6.19* 10^-3m

 

6 0
3 years ago
A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rat
Inessa05 [86]

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s

4 0
3 years ago
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density
blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

6 0
3 years ago
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