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2 years ago

6th grade science please help!

Physics

1 answer:

labwork [276]2 years ago

5 0

Answer:

conduction

Explanation:

Conduction transfers heat via direct molecular collision. An area of greater kinetic energy will transfer thermal energy to an area with lower kinetic energy. Higher-speed particles will collide with slower speed particles.

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If she asks you to cut a strip of pie crust long around the outside of the pan, how long does it need to be? (Remember c/d =3.14

Nookie1986 [14]

I remember c/d. That's not a problem. But if you want 'c', you'll have to give me 'd'.

5 0

3 years ago

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

3 years ago

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

centripetal force acting on satellite = $\dfrac{mv^2}{r}$

gravitational force = $\dfrac{GMm}{r^2}$

equating both the above equation

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

$v = \sqrt{\dfrac{GM}{r}}$

$v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

T = 14416.92 s

$T = \dfrac{14416.92}{3600}\ hr$

T = 4 hr

c) gravitational force acting

$F = \dfrac{GMm}{r^2}$

$F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

F = 5202 N

4 0

3 years ago

The period of a sound wave coming from an instrument is 0. 002 seconds. What is the frequency of the sound? Hz.

Pie

The period is the time taken by the wave to complete an oscillation. The frequency of the given sound is 500 Hz.

<h2>Period:</h2>

It is the time taken by the wave to complete an oscillation. The frequency is inversely proportional to the time:

$f = \dfrac 1T$

Where,

$f$ - frequency

$T$ - period = 0.002 s

Put the value in the equation,

$f = \dfrac 1{0.002}\\\\f = 500\rm \ Hz$

Therefore, the frequency of the given sound is 500 Hz.

Learn more about Period:

brainly.com/question/842349

5 0

2 years ago

If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force betw

LiRa [457]

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

4 0

3 years ago

6th grade science please help!​

6th grade science please help!