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Romashka [77]
2 years ago
11

The following coplanar forces pull on a ring 200N at 30 degrees and 500N at 80 degrees and 300N at 240 degrees and an unknown fo

rce. Find the magnitude and direction of the unknown force of the ring is to be in equilibrium
Physics
1 answer:
Alex73 [517]2 years ago
3 0

MAGNITUDE AND DIRECTION OF A VECTOR

Given a position vector →v=⟨a,b⟩,the magnitude is located by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba)

<h3>What is the formula of magnitude?</h3>

The formula to determine the extent of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the procedure to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)

To learn more about magnitude and direction, refer

brainly.com/question/27248532

#SPJ9

main answer is given below,

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In which one of the following situations is zero net work done? A) A ball rolls down an inclined plane. B) A physics student str
lidiya [134]

Answer:

Option D

Explanation:

The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.

In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.  

In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

6 0
4 years ago
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When you drop a 0.43 kg apple, Earth ex erts a force on it that accelerates it at 9.8m / (s ^ 2) toward the earth's surface. Acc
Sever21 [200]

Answer:

The answer to the question is highlighted in the box

8 0
1 year ago
A normal human heart, beating about once per second, creates a maximum 4.00 mV 4.00 mV potential across 0.250 m 0.250 m of a per
ElenaW [278]

Answer:

1) maximum electric field strength = 0.016 V/m

2)maximum magnetic field strength = 5.33 x 10^(-11) T

3) wavelength = 3 x 10^(8)m

Explanation:

A) We are given;

Frequency (f) = 1 Hz

Maximum Potential ΔV = 4 mV = 4 x 10^(-3) V

Length; d = 0.25 m

Now, maximum potential formula is given as;

ΔV = E_o•d

Where E_o is maximum electric field strength.

Thus,

E_o= ΔV/d = (4 x 10^(-3))/0.25 = 0.016 V/m

B) The maximum magnetic field strength is given by;

B_o = E_o/c

Where;

c is speed of light = 3 x 10^(8)

B_o is maximum magnetic field strength

Thus,

B_o = 0.016/(3 x 10^(8)) = 5.33 x 10^(-11) T

C) Speed of any electromagnetic wave is given by;

c = fλ

Where;

f is frequency

λ is wavelength

c is speed of light = 3 x 10^(8)

Thus, making λ the subject,

λ = c/f

λ = (3 x 10^(8))/1 = 3 x 10^(8)m

8 0
4 years ago
Energy is always conserved. It cannot be created or destroyed. It can, however, be transferred between objects or systems, from
sweet-ann [11.9K]

Have in mind: Energy is always conserved and, (...) be transferred between objects or systems, from one form to another.

Hence, the true relation is C, when a system gains kinetic energy, it loses potential energy.

Think about this,

K + P = constant

If K increases, then P need to decreases to holds the above relation.

5 0
3 years ago
A projectile of mass 6.8 kg kg is shot horizontally with an initial speed of 14.5 m/s from a height of 26.7 m above a flat deser
sleet_krkn [62]

Answer:

Explanation:

Given

mass of projectile m=6.8\ kg

initial horizontal speed u_x=14.5\ m/s

height h=26.7\ m

Considering vertical motion

velocity gained by projectile during 26.7 m motion

v^2-u^2=2 as

v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-(0)^2=2\times (9.8)\times (26.7)

v=\sqrt{523.32}

v=22.87\ m/s

Horizontal velocity will remain same as there is no acceleration

final velocity v_{net}=\sqrt{(v)^2+(u_x)^2}

v_{net}=\sqrt{733.57}=27.08\ m/s

Initial kinetic Energy K_i=\frac{1}{2}mu_x^2

K_i=\frac{1}{2}\times 6.8\times (14.5)^2=714.85\ J

Final Kinetic Energy K_f=\frac{1}{2}mv_{net}^2

K_f=\frac{1}{2}\times 6.8\times (27.08)^2

K_f=2493.30\ J

Work done by all the force is equal to change in kinetic Energy of object

Work done by gravity is W_g

W_g=\Delta K

W_g=2493.30-714.85=1778.45\ J  

8 0
3 years ago
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