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3 years ago

HURRY PLEASE ILL MARK YOU AS BRAINLIEST! 30 POINTS!

Physics

2 answers:

saveliy_v [14]3 years ago

6 0

The correct answer is B .

Electrically charged particles

there i gave you another person answer

kap26 [50]3 years ago

5 0

The correct answer is B .

Electrically charged particles.

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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

3 years ago

•. Parking orbit is -

notsponge [240]

Answer:

Parking orbit is -

a. The path along which a plane travels

b. Orbit of a polar satellite

c. Orbit of geostationary satellite

d. Orbit of the earth.

Explanation:

7 0

3 years ago

Read 2 more answers

A wave traveling in water has a frequency of 500.0 Hz and a wavelength of 3.00 m. What is the speed of the wave?

Sergeu [11.5K]

Answer:

1500 m/s

Explanation:

Recall that for a wave,

Speed = frequency x wavelength

here we are given frequency = 500 Hz and wavelength = 3m

simply substitute into above equation

Speed = 500 Hz x 3m

= 1500 m/s

6 0

3 years ago

An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.

rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = $\frac{1}{2}$ $m_{p}$ v²

v = √( 2K / $m_{p}$ )

lets relate the cross-sectional area A of the beam to its diameter D;

A = $\frac{1}{4}$ πD²

now, we substitute for v and A

n = I / $\frac{1}{4}$ πeD² ×√( 2K / $m_{p}$ )

n = 4I/π eD² × √( $m_{p}$ / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵

n = 3.2 × 10¹³ m⁻³

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0

2 years ago

Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical

inysia [295]

Answer:

Gas 2, Gas 3, Gas 4, Gas 5 is the order of decreasing strength of inter-molecular forces.

Explanation:

The strength increases as there is a decrease in the vanderwaals constant and vice versa.

3 0

3 years ago