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yarga [219]
3 years ago
7

I need help in physics please help me

Physics
1 answer:
artcher [175]3 years ago
3 0

The first step is to represent the vectors shown in the image in Cartesian coordinates.

For the vector C we have a magnitude of 4.8 and an angle 22 ° with the axis -y (direction j)

To write this vector in Cartesian coordinates we must find its component in x (address i) and in the y axis.

x = 4.8sin(22)i\\\\y = 4.8cos(22)(-j)

So:

C = 4.8sin(22) i + 4.8cos(22)(-j)\\\\C = 1.798\ i - 4.450\ j

For Vector B we have a magnitude of 5.6 and an angle of 33 with the -x axis (-i direction)

So:

x = 5.6cos(33)(-i)\\\\y = 5.6 sin(33)(-j)

So:

B = 5.6cos(33)(-i) + 5.6sin(33)(-j)\\\\B = -4.696\ i - 3.05\ j

Finally the sum of B + C is made component by component in the following way:

F = (-4.696 +1.798)i + (-4.450 - 3.05)j\\\\F = -2.898\ i - 7.5\ j

Finally the magnitude of f is:

|F| = \sqrt{(-2,898)^2 + (-7.5)^2}

| F | = 8.04

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3 years ago
You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator
ICE Princess25 [194]

Answer:

A) mass = 3121.58 kg

B) tension = 25940.37 N

C) tension = 25940.37 N (tension on both sides will be the same)

Explanation:

Weight of elevator = 22500 N

Distance = 6.75 m

Time = 3 sec

Since it started from rest, initial speed is zero.

Using Newton's equation of motion we have,

S = ut + 0.5at^2

S = distance covered = 6.75 m

t = time = 3 s

a = acceleration upwards

u = initial velocity = 0

Substituting values, we have,

6.75 = 0(3) + (0.5 x a x 3^2)

6.75 = 4.5a

a = 6.75/4.5 = 1.5 m/s^2 (acceleration of the elevator upwards)

Mass of the elevator = Weight/g

Where g = acceleration due to gravity 9.81 m/s

Mass = 22500/9.81 = 2293.58 kg

From the image below we solve from

T - 22500 = ma

T - 22500 = 2293.58 x 1.5

T - 22500 = 3440.37

T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)

On the other side,

mg - T = ma

9.81m - 25940.37 = 1.5m

(9.81 - 1.5)m = 25940.37

8.31m = 25940.37

m = 3121.58 kg (mass of counter weight)

See image below

8 0
3 years ago
An electric field is produced by the very long, uniformly charged rod drawn above. If the strength of the electric field is E1 a
Debora [2.8K]

Answer: hello the complete question is attached below

answer :

r2 = 4r1

Explanation:

Electric field strength = F / q

we will assume the rod has an infinite length

For an infinitely charged rod

E ∝ 1/ r

considering two electric fields E1 and E2 at two different locations as described in the question

E1/E2 = r1/r2 ----- ( 2 )

<u>Calculate for r2 when E2 = E1/4 </u>

back to equation 2

E1 / (E1/4) = r1 / r2

∴ r2 = 4r1

3 0
3 years ago
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