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3 years ago

Name the missing force... Normal Spring Tension Air Resistance

Physics

1 answer:

Kay [80]3 years ago

4 0

the missing force is spring force.

The object is hanging from the spring and the spring is stretched by some distance from its equilibrium position. due to this stretch in the spring , a spring force starts acting on the object trying to regain its equilibrium position.

the spring force is given as

F = kx

where F = spring force ,k = spring constant , x = stretch in the spring.

the spring force balances the weight of the object in down direction and hence keeps the block from falling down.

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Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho

Yuri [45]

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

$F_{A} + F_{B} + F_{C} = 0$ (1)

Si sabemos que $F_{A} = -3$ y $F_{B} = 5$ , entonces el valor de la fuerza que debe ejercer la persona C debe ser:

$F_{C} = -F_{A}-F_{B}$

$F_{C} = -(-3)-5$

$F_{C} = -2$

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

3 0

3 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago

Explain in your own words how the Doppler Effect is also applicable in our study of light.

arlik [135]

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

8 0

3 years ago

Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates

IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

Sphere A : ma = 80 kg , ( 0 , 0 )

Sphere B : ma = 60 kg , ( 0.25 , 0 )

Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

Determine the angle (α) between vectors rac and rab using cosine rule:

$cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}$

Determine the angle (β) between vectors rbc and rab using cosine rule:

$cos ( \beta ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta ) = 0.6\\\\\beta = 53.13^{\circ \:}$

- Now determine the scalar gravitational forces due to sphere A and B on C:

Between sphere A and C:

Fac = G*ma*mc / rac^2

Fac = (6.674×10−11)*80*0.2 / 0.2^2

Fac = 2.67*10^-8 N

vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

Between sphere B and C:

Fbc = G*mb*mc / rbc^2

Fbc = (6.674×10−11)*60*0.2 / 0.15^2

Fbc = 3.56*10^-8 N

vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

Fc = vector Fac + vector Fbc

Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8

Fc = [ - 4.45 * 10^-8 j ] N

3 0

3 years ago

GETS BRAINILIST!!!!Which of the following will increase the gravitational force exerted by one object on another? Increasing the

jeka94

Answer:

C. Decreasing the distance between the objects.

Explanation:

Gravitational force depends on mass and distance. The farther the objects are apart, the less gravitational force/pull they have.

5 0

3 years ago