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3 years ago

Forces that act on an object but are not equal in size

1 answer:

7 0

Forces that are equal in size but opposite in direction and do not cause a change in an object's movement are called balanced forces.

forces that aren't equal in size and do cause a change in movement (what it seems like you're asking for) are called UNBALANCED FORCES

so answer (in case that wasn't clear, as I'm tired) : unbalanced forces

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What is the mathematical relationship between wavelength and energy transmission?.

Tomtit [17]

Answer: Wavelength is related to light and also related to energy. The shorter the wavelengths and the higher the frequency corresponds with greater energy. So the longer the wavelengths and lower the frequency results in lower energy. The energy equation is E = hν.

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2 years ago

Qué ocurre con el espectro cuando se interpone el celofán de colores?

Travka [436]

Answer: You will only see the color that cellophane lets through

Explanation:

Let's begin by the fact the whole electromagnetic spectrum is known as "white light", which is composed by a range of colors (wavelengths).

Now, if we have a source with white light (the Sun, for example) and we interpose a cellophane of any color (let's choose red), this cellophane will act as a filter and will only let pass the color of the cellophane.

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3 years ago

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

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3 years ago

11. Which of the following is a proper unit of

Sergeeva-Olga [200]

D is the correct answer!!

4 0

3 years ago

Read 2 more answers

a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between t

laiz [17]

Answer: A

Explanation:

From the question, the given parameters are given.

Mass M = 30kg

Radius r = 2 m

Coefficient of static friction μ = 0.8

Coefficient of kinetic friction μ = 0.6

Kinetic friction Fk = μ × mg

Fk = 0.6 × 30 × 9.8

Fk = 176.4 N

The force acting on the merry go round is a centripetal force F.

F = MV^2/r

This force must be greater than or equal to the kinetic friction Fk. That is,

F = Fk

F = 176.4

Substitute F , M and r into the centripetal force formula above

176.4 = (30×V^2)/2

Cross multiply

352.8 = 30V^2

V^2 = 352.8/30

V = sqrt (11.76) m/s

V = 5.24 m/s

Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately

4 0

3 years ago