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3 years ago

8

How can different objects interact when they are not even touching?

1 answer:

jeka943 years ago

3 0

Answer: Magnetic and gravitational force

Explanation: When a magnet and an iron nail are kept at a distance,the magnet attracts the nail without touching using magnetic force. In this example, the magnet and the nail are interacting.

The earth pulls the moon towards it and keeps it in orbit without touching it, using gravitational force. In this example,the moon and the earth are interacting.

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A ticker-tape is moved through a ticker-timer for 5 seconds . if the timer is operating at 25 Hz. i.How many dots would have bee

svetlana [45]

Answer:

Explanation:

25 cycles/s (5 s) = 125 dots

Tape does Repetitious motion.

It might possibly be harmonic motion, but we do not have enough information to say for certain.

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2 years ago

Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$ F be the temperature and T that of the body

Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = <em>kdt</em>

In [T-70]= <em>kt </em>+ $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for <em>k </em>using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

<em>K </em>= $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230 $e^{-0.19018}$

4 0

3 years ago

Box A is pulled 2 meters when a 3-newton force is applied to it.

murzikaleks [220]

Answer: 0

Explanation: 1-1

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3 years ago

Jasper and Gemma are going to play on a teeter totter. Gemma gets on first. When Jasper gets on, Gemma moves into the air, but s

Nezavi [6.7K]

Answer:

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Explanation:

7 0

3 years ago

Read 2 more answers

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

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

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

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