All categories

3 years ago

Find the magnitude of the resultant of forces 6N and 8N acting at 240° to each other

Physics

1 answer:

Alexxx [7]3 years ago

6 0

Answer:

magnitude of the resultant of forces is 11.45 N

Explanation:

given data

force F1 = 6N

force F2 = 8N

angle = 240°

solution

we get here resultant force that is express as

F(r) = $\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}$ ..............1

put here value and we get

F(r) = $\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}$

F(r) = 11.45 N

so magnitude of the resultant of forces is 11.45 N

You might be interested in

Please help i need it please

Ronch [10]

ANSWER:
Ca(OH)2 shows a decrease of solubility with the increase of temperature.

Hope it helps u! :)

6 0

3 years ago

Read 2 more answers

The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

n200080 [17]

<h2>Answer:g=9.79 $ms^{-2}$ ,A object of mass $m$ at the surface of earth experiences a force $mg$ </h2>

Explanation:

Let $M$ be the mass of earth.

Let $R$ be the radius of earth.

Let $G$ be the universal gravitational constant.

Given,

$M=5.96\times 10^{24}Kg$

$R=6.37\times 10^{6}m$

$G=6.67259 \times 10^{-11}$ $Nm^{2}Kg^{-2}$

Let $g$ be the acceleration due to gravity.

Then, $g=\dfrac{GM}{R^{2}}$

$g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}$

$g=9.79ms^{-2}$

A object of mass $m$ at the surface of earth experiences a force $mg$

3 0

3 years ago

A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

3 years ago

sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coeﬃcient

lara31 [8.8K]

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

6 0

4 years ago

A ball is thrown vertically upward with a speed v. an identical second ball is thrown upward with a speed 2v. how many times hig

vladimir2022 [97]

It goes twice as fast as the first one. Twice

6 0

3 years ago