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soldi70 [24.7K]
2 years ago
9

Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni

tude of the resultant vector A + B be?
Physics
1 answer:
Fudgin [204]2 years ago
3 0

Answer:

|\vec A + \vec B| = 50 units

Explanation:

As we know that magnitude of two vectors is given as

|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}

here we know that

A = magnitude of vector A

B = magnitude of vector B

\theta = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}

|\vec A + \vec B| = \sqrt{30^2 + 40^2}

|\vec A + \vec B| = 50 units

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disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

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d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

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2 years ago
Two identical stones are thrown from the top of a tall building. Stone 1 is thrown vertically downward with an initial speed v,
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If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by

v= \frac{1}{2}gt

Where

g = Gravitational acceleration

t = time

As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.

Both bodies will end with the same thermal speed.

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2 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

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I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

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Answer:

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Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

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O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

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Answer

32000

Explanation:

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