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3 years ago

When parts of your body functions as lever systems, what part supplies the force?

1 answer:

5 0

Muscles supply the force. They exert force by getting shorter (contracting).
A muscle can't get longer by itself. There needs to be another muscle
pulling in the opposite direction.

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10. The electron dot diagram for the element Ne would have

zmey [24]

Answer:

2,8

Explanation:

The first electron shell would have 2 electrons, the second shell would have 8 electrons. This is because Neon has a relative charge of 10.

7 0

2 years ago

True or False:<br><br> It is safe to touch an electric current while soaking wet.

olganol [36]

Answer:

False

Explanation:

Water contains ions that can conduct electricity and if touched it can cause harm (aka electrocution).

8 0

2 years ago

When ultra violets lights shine on glass what does it do to electrons in the glass structure?

andreev551 [17]

Answer:

Explanation:

7 0

2 years ago

A train travels 200km/hr. how much distance will the train be from the station in 2.5 hours?

SCORPION-xisa [38]

Answer:

500km

Explanation:

Given parameters:

Speed = 200km/hr

Time taken = 2.5hrs

Unknown:

Distance = ?

Solution:

To solve this problem, we use the speed, time and distance equation.

Therefore;

Distance = Speed x time

So;

Distance = 200 x 2.5 = 500km

6 0

2 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

2 years ago