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2 years ago

12

Chang wants to build a circuit that will light up the bulb from a flashlight. Which component could Chang leave out and still li

ght up the bulb? the switch the battery the wire the light bulb

2 answers:

goldfiish [28.3K]2 years ago

7 0

Answer:

A) The Switch

Explanation:

I took the test and got it right :) goodluck!

Elan Coil [88]2 years ago

4 0

The component that could Chang leave out and still light up the bulb is the switch.

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What an image of a distant object is brought into focus in front of a persons retina the fact is called

beks73 [17]

When the image of a distant object is brought into focus of front of a person's retina, the defect is called: nearsightedness.

7 0

3 years ago

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

3 years ago

The moon's surface gravity is one-sixth that of the earth. Calculate the weight on the moon of an object that has a mass of 24 k

ad-work [718]

When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
we mean that the acceleration of gravity on the Moon's surface is 1/6 of
the acceleration of gravity on the Earth's surface.

The acceleration of gravity is (9.8 m/s</span>²) on the Earth's surface, so
<span>it would be (9.8/6 m/s</span>²) on the Moon's surface.
<span>
The weight of any object, right now, is

(object's mass) </span>· (acceleration of gravity where the object is located now) .
<span>
If the object's mass is 24 kg and the object is on the Moon right now,
then its weight is

(24 kg) </span>· (9.8/6 m/s²)

= (24 · 9.8 / 6) kg-m/s²

= 39.2 Newtons

7 0

3 years ago

A falcon can descend with a speed 250 km/h. If a falcon flies at this speed for 2.0 s and then flies a 100 m in 2.5 s, what is t

Vadim26 [7]

Answer:

v= s/t

Explanation:

250 km/ h =69.44m/s

S1=2 times 69.44 ≈ 139m

Next 2.5 seconds:

S2 = 100m

Average speed:

v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h

3 0

2 years ago

Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q

katen-ka-za [31]

Answer:

c) $4.2*10^{-5}C$

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

$F=\frac{kq_{1}q_{2}}{r^{2}}$

where k is a proportionality constant with the value $k=9*10^{9}\frac{Nm^{2}}{C^{2}}$

In this case $q_{1}=q_{2}=q$ , so we have:

$F=\frac{kq^{2}}{r^{2}}$

Solving the equation for q, we have:

$kq^{2}=Fr^{2}$

$q^{2}=\frac{Fr^{2}}{k}$

$q=\sqrt{\frac{Fr^{2}}{k}}$

Replacing the given values:

$q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}$

$q=4.2*10^{-5}C$

3 0

3 years ago