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3 years ago

Explain why some objects fall faster than other objects,

1 answer:

8 0

objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.

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If i try to fail and succeed which one did I do

Novay_Z [31]

You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.

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3 years ago

A man has 2 spheres A and B. He gently drops sphere A vertically down and throws the sphere B horizontally at the same time. Whi

anyanavicka [17]

Answer:

The answer is c

Explanation:

5 0

2 years ago

Friction depends on the types of surfaces involved and how hard the surfaces push together. Please select the best answer from t

LenKa [72]

True: Friction depends on the types of surfaces involved and how hard the surfaces push together.

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3 years ago

he Hobbits are building a watchtower so they can prepare to battle in case trolls decide to attack them. One Hobbit will always

joja [24]

Answer:

5m/8

Explanation:

Function T gives the time the Hobbits have to prepare for the attack, T(k), in minutes, as a function of troll's distance, k, in meters.

Function V gives visibility from the watchtower, V(m), in meters, as a function of the height of the watchtower, m, in meters.

Therefore, T(V(m)) will give the time the Hobbits have to prepare for the troll attack as a function of the height, m, of the watchtower.

We can input m into function V to obtain the visibility from watchtower, V(m), in meters. Since visibility indicates the distance you can see, this also gives the distance of the trolls. This can then be input into function T to obtain the time that the Hobbits have to prepare for a troll attack.

Let's find T(V(m)) by substituting the formula for V(m) into function T as shown below.

T(V(M))=T(50m)

=50m/80

We can simplify this as follows:

=50m/80

=5m/8

4 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago