All categories

2 years ago

Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.

1 answer:

7 0

100cos30° + 80cos120° + 40cos233° + Dx = 0
Dx = -22.53 N 

y components: 
100sin30° + 80sin120° + 40sin233° + Dy = 0 
Dy = -87.34 N 

magnitude of D: 
sqrt[(-22.53)² + (-87.34)²] 
90.2 N 

direction of D: 
arctan[(-87.34)/(-22.53)] 
75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: 
360 - 75.5° = 284.5°

You might be interested in

The maximum mass of a white dwarf is ______ times the mass of the sun.

Savatey [412]

Answer:

(A)

Explanation:

about 1.4 times the mass of our sun.

8 0

2 years ago

Particle of the nucleus that has no electrical charge

Helga [31]

That would be the atom I believe.

3 0

3 years ago

Read 2 more answers

A ball rolls off a table with a horizontal velocity of 3 m/s. If it takes 0.3 seconds for the ball to reach the floor, how high

LekaFEV [45]

The height of the table above the ground is 0.45 m.

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Horizontal velocity (u) = 3 m/s
Time (t) = 0.3 s
Acceleration due to gravity (g) = 10 m/s²
Height (h) =?

<h3>How to determine the height </h3>

The height of the table can be obtained by using the following formula:

h = ½gt²

h = ½ × 10 × 0.3²

h = 5 × 0.09

h = 0.45 m

Thus, the height of the table is 0.45 m

Learn more about motion under gravity:

brainly.com/question/26275209

6 0

2 years ago

In an electromagnetics lab, you are studying two coils which have a mutual inductance of M=300 mH. Suppose that the current in t

Firdavs [7]

Answer:

Explanation:

Given that,

The mutual inductance of the two coils is

M = 300mH = 300 × 10^-3 H

M = 0.3 H

Current increase in the coil from 2.8A to 10A

∆I = I_2 - I_1 = 10 - 2.8

∆I = 7.2 A

Within the time 300ms

t = 300ms = 300 × 10^-3

t = 0.3s

Second Coil resistance

R_2 = 0.4 ohms

We want to find the current in the second coil,

The same induced EMF is in both coils, so let find the EMF,

From faradays law

ε = Mdi/dt

ε = M•∆I / ∆t

ε = 0.3 × 7.2 / 0.3

ε = 7.2 Volts

Now, this is the voltage across both coils,

Applying ohms law to the second coil, V=IR

ε = I_2•R_2

0.72 = I_2 • 0.4

I_2 = 0.72 / 0.4

I_2 = 1.8 Amps

The current in the second coil is 1.8A

7 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago