All categories

1 year ago

two forces 3N and 4N act on a body in a direction due north and due East respectively calculate their equivalent

1 answer:

6 0

Two forces 3N and 4N act on a body in a direction due north From East, the equilibrant's angle is given by $\theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}$ .

<h3>What are equilibrium and resultant force?</h3>

The equilibrium force is the balanced force when the net force acting is zero and is the exact opposite of the consequent force. The resultant force is one single force replaced by numerous forces.

<h3>Briefing:</h3>

3N and 4N are the two forces pulling on a body.

The forces work along the North and the East, which are perpendicular to one another.

The resultant of the forces, which is provided by the equilibrant force,

R = √(3)²+(4)²

R = 5N

From East, the equilibrant's angle is given by

$\theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}$

To know more about equilibrium force visit:

brainly.com/question/12582625

#SPJ9

You might be interested in

Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laborat

meriva

Answer:

k = 200 N/m

Explanation:

given,

mass of the object = 251 g

= 251 x 10⁻³ Kg = 0.251 Kg

distance of tendon stretch = x = 1.23 cm

= 1.23 x 10⁻² = 0.0123 m

using the formula

F = k x

where

k is the force constant of the tendon

F = m g

F = 0.251 x 9.8 = 2.4598 N

2.4598 = k x 0.0123

k = 199.98 N/m

k = 200 N/m

hence, force constant of the tendon is approximately equal to 200 N/m

5 0

3 years ago

The amplitude, or magnitude, of a sinusoidal source is the maximum value of the source. What is the amplitude of the voltage sou

liraira [26]

Answer:

Explanation:

<u>Sinusoidal Functions</u>

A sinusoid or sinusoidal function is a sine or cosine which general equation is

$F(x)=A.sin(wt-\phi)$

Or also

$F(x)=A.cos(wt-\phi)$

Where A is the amplitude or maximum value, w is the angular frequency, t is the time and $\phi$ is the phase shift.

Comparing the given expression with the general formula

$v(t)=50cos(2000t-45^o) mV$

We can establish that A=50 mV = 0.05 V

$A = 0.05\ V$

7 0

3 years ago

A monochromatic light passes through a narrow slit and forms a diffraction pattern on a screen behind the slit. As the slit widt

maw [93]

Answer:

shrinks with all the fringes getting narrower

Explanation:

As the light passes through the slit, the diffraction pattern shrinks, as the waves have more opening to penetrate, and the fringes becomes more narrow as a result of that, The opposite happens as the conditions are reversed.

8 0

3 years ago

The contribution of Tycho Brahe was primarily his

mihalych1998 [28]

Tycho Brahe ( 1546 - 1601 ) was a Danish astronomer known for his accurate astronomical and planetary observations. Tycho tried to produce a model with the best of both Ptolemy ( earth-centered solar system ) and Copernicus ( sun-centered solar system ).
Answer: B ) observation.

3 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago

two forces 3N and 4N act on a body in a direction due north and due East respectively calculate their equivalent​

two forces 3N and 4N act on a body in a direction due north and due East respectively calculate their equivalent