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ivolga24 [154]
3 years ago
5

Some rearview mirrors produce images of cars to your rear that are smaller than they would be if the mirror were flat.Are the mi

rrors concave or convex? What is a mirrors radius of curvature if cars 20.8 m away appear 0.33x their normal size?
Physics
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

Concave mirror,-20.48m

Explanation:

Concave mirrors-This is because the objects are made smaller by the mirror, so they appear farther away.

#we find the image distance from the magnification:

m=\frac{h_1}{h_o}=\frac{-d_i}{d_o}\\\\+0.33=\frac{-d_i}{20.8}\\\\d_i=-6.864m

#We find the focal length from:

({\frac{1}{d_o})}+({\frac{1}{d_i})}=({\frac{1}{f})}\\\\({\frac{1}{20.8})}+({\frac{1}{-6.864})}=\frac{1}{f}\\\\f=-10.24m

r=2f=-10.24m*2=-20.48m

#The mirror has a negative radius(-20.48m) of curvature since its vertex  lies to the right of the center of curvature

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Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

F = qv \times B = qvBsin(\theta)     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

K = \frac{1}{2}mv^{2}

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

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a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

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Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

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Thus, the velocity after collision is 11.456 m/s.

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Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

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