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Anna [14]
2 years ago
15

Light from a distant star is collected by a concave mirror that has a radius of curvature of 150 cm. How far from the mirror is

the image of the star?
Physics
1 answer:
andrey2020 [161]2 years ago
4 0
The focal point of a mirror is half of the radius of curvature. We can use the formula,                                                                                                                   1/f = 1/v + 1/u                                                                                                          where f is the focal length , v is the image distance and u is object distance        The distance of the star is assumed to be incredibly far away and 1 divided by a really big number is approximately zero, thus;                                                     1/f = 1/v  = 1/75 = 1/v                                                                                             therefore; the image is formed 75 cm infront of the mirror
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Upon impact, bicycle helmets compress, thus lowering the potentially dangerous acceleration experienced by the head. A new kind
Dimas [21]

Answer:

acceleration = -15.3g

Explanation:

given data

speed = 6.00 m/s.

thickness = 12

moves the entire = 12.0 cm

solution

we will use here equation that is

v² - u²  = 2 × a × s    ........................1

here v = 0 is the final velocity and u = 6.0 m/s is initial velocity and s= 0.12 m is the distance covered and a is the acceleration

so we put here value and get acceleration

a = \frac{v^2-u^2}{2s}

a = \frac{0^2-6^2}{2\times 0.12}

a = -150 m/s² ( negative sign means it is a deceleration )

and

acceleration in units of g  

a = \frac{-150}{9.8}

a = -15.3 g

6 0
2 years ago
Describe electrons.<br> Location:<br> Charge:<br> Mass
irga5000 [103]
Protons, neutrons and electrons. Both protons and neutrons have a mass of 1 amu and are found in the nucleus. However, protons have a charge of +1, and neutrons are uncharged. Electrons have a mass of approximately 0 amu, orbit the the nucleus, and have a charge of -1.
5 0
2 years ago
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Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
Advocard [28]

Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax =  27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16  sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

   = √27²+16² - (2(27)(16)(cos 60))

   = 23.52 m

now we can use the sines law to find the angle

tan θ = \frac{R_{y} }{R_{x} }

         = By - Ay/Bx -Ax

         = (-8 - 13.5)/(-13.9 - (-23.4))

     θ  = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

         = 24° east of south

4 0
2 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
balandron [24]

Answer with Explanation:

We are given that

Radius of  solid core wire=r=2.28 mm=2.28\times 10^{-3} m

1mm=10^{-3} m

Radius of each strand  of thin wire=r'=0.456 mm=0.456\times 10^{-3} m

Current density of each wire=J=3750 A/m^2

a.Area =\pi r^2

Where \pi=3.14

Using the formula

Cross section area of copper wire has solid core =3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2

Current density =J=\frac{I}{A}

Using the formula

3750=\frac{I}{16.3\times 10^{-6}}

I=3750\times 16.3\times 10^{-6}=0.061 A

Total number of strands=19

Area of strand wire=A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2

J'=\frac{I'}{A'}

3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}

I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2

I'=0.047 A

b.Resistivity of copper wire=\rho=1.69\times 10^{-8}\Omega-m

Length of each wire =6.25 m

Resistance, R=\frac{\rho l}{A}

Using the formula

Resistance of solid core wire=R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega

Resistance of strand wire=R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega

7 0
2 years ago
A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a
Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=mv_{2} +mv_{1}

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0
2 years ago
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