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BlackZzzverrR [31]
2 years ago
6

vector A makes equal angles with x,y and z axis. value of its components (in terms of magnitude of vector A will be?

Physics
2 answers:
kiruha [24]2 years ago
5 0
X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared
djverab [1.8K]2 years ago
5 0

Answer:

All three components are

A_x = \frac{A}{\sqrt3}

A_y = \frac{A}{\sqrt3}

A_z = \frac{A}{\sqrt3}

Explanation:

As we know that sum of all three components of the vector will give us resultant vector

So here we can say that

A^2 = A_x^2 + A_y^2 + A_z^2

since it is given that all three components are of same magnitudes so

A_x = A_y = A_z

now we have

A^2 = 3A_x^2

so we have

A_x = A_y = A_z = \frac{A}{\sqrt3}

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The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b
Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0
2 years ago
Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown
olasank [31]

Answer:

a)  v = √ g x , b)  W = 2 m g d , c)    a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

            T-W₁ = m a

            W₃ - T = M a

            w₃ - w₁ = (m + M) a

            a = (3m - m) / (m + 3m) g

            a = 2/4 g

            a = ½ g

the speed of the blocks is

          v² = v₀² + 2 ½ g x

          v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

           W₁ = -W d = - mg d

3m heavy block

             

            W₂ = W d = 3m g d

the total work is

             W = W₁ + W₂

             W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

8 0
2 years ago
Read 2 more answers
To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj
ExtremeBDS [4]

Answer:

1.138\times 10^{-3}V

Explanation:

Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:

\epsilon=\frac{d\phi}{dt}

where:

d\Phi-variation of the magnetic flux

dt-is the variation of time

#The magnetic flux through the coil is expressed as:

\Phi=NBA \ Cos \theta

Where:

N- number of circular loops

A-is the Area of each loop(A=\pi r^2=\pi \times 5^2=78.5398)

B-is the magnetic strength of the field.

\theta=15\textdegree- is the angle between the direction of the magnetic field and the normal to the area of the coil.

\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}

\frac{dB}{dt}-=0.0250T/s is given as rate at which the magnetic field increases.

#Substitute in the emf equation:

=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V

Hence, the induced emf is 1.138\times 10^{-3}V

4 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
Seven seconds after a brilliant flash of lightning, thunder shakes the house. approximately how far was the lightning strike fro
tangare [24]
Very roughly 7,700 feet ... about 1.5 miles.
8 0
2 years ago
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