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Vladimir [108]
3 years ago
9

If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced

?
A.
real, upright, and magnified
B.
virtual, inverted, and smaller
C.
virtual, upright, and magnified
D.
real, inverted, and magnified
E.
real, upright, and smaller
Physics
1 answer:
pshichka [43]3 years ago
3 0

Answer: I think its D

Explanation: Hope this was helpful...

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The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object i
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Answer:

Explanation:

1) Force Friction = Normal Force * Coefficient of Friction

Force Friction = Mass * Gravity * Coefficient of Friction

2) F = ma

Force = mass * acceleration

Force Friction (from #1) = mass * acceleration

acceleration = Force Friction / Mass

6 0
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a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

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7 0
3 years ago
It requires a force of 100 N to stretch a spring of negligible mass by a distance of 0.1 m. If the spring instead stretches a di
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by the formula of spring force we know that

F = kx

here we know that

F = 100 N

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force is given by

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5 0
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Read 2 more answers
A person holds a rifle horizontally and fires at a target. The bullet leaves the muzzle of the rifle with a velocity of 460 m/s.
Trava [24]

Answer:

the distance travelled from the bullet to the target  is 391m

Explanation:

Hello! To solve this exercise we must follow the following steps.

1. the bullet travels with constant speed which means that the distance traveled to the target is given by the following equation

X=(V1)(T1)

T1=\frac{X}{V1} =\frac{x}{460}

where

X=target distance

V1=bullet speed=460m/s

T1=

time it takes for the bullet to reach the target

2. The distance the sound travels is given by the following equation (it is the same as the distance from the person to the target)

X=(V2)(T2)

T2=\frac{X}{V2} =\frac{x}{340}

X=

target distance

V2= speed of sound=340m/s

T2=   time it takes the sound of the Bullet to return.

3. The total time it takes for the person to hear the bullet(T=2s) is the sum of the time it takes for the bullet to reach the target, plus the time it takes for the sound to reach the person, with the above we infer the following equation

T=T1+T2

2=T1+T2

4. Finally we use the equations found in step 1 and 2 to find the distance traveled using algebra.

2=\frac{x}{340}+\frac{x}{460} \\x(\frac{1}{340} +\frac{1}{460} )=2\\\ X= \frac{2}{(\frac{1}{340} +\frac{1}{460} )} \\\\x=391m

the distance travelled from the bullet to the target  is 391m

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3 years ago
A high frequency sound wave corresponds to high _____.
lina2011 [118]
A highly frequency sound wave corresponds to a high pitch sound
6 0
3 years ago
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