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Paladinen [302]
3 years ago
7

Dierdre drew a diagram to compare the three types of mirrors.

Physics
1 answer:
Sholpan [36]3 years ago
8 0

<em>Labels that belong in the marked ares X, Y & Z include;</em>

X: Curves outward

Y: Image may be smaller than object

Z: Image is always virtual

<u>Since the rays never meet, the images formed by convex mirrors are always virtual and smaller than the object, and since they are smaller, the images appear to be further than they actually are.</u>

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There are 2.2 pounds in a kilogram. if a boys mass is 40kg, what is his height in pounds?
sasho [114]
The simplest way to do this is to set up equivalent fractions, like this- 

\frac{1}{2.2} = \frac{40}{x}

Solve for x by using cross multiplication.

40*2.2= 88
1*x=88
x=88

Therefore, the boy weighs 88lbs. 
3 0
3 years ago
Read 2 more answers
Pls how to solve this problem. help would be appreciated ​
Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0
3 years ago
A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-1
gladu [14]

Answer:

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert
strojnjashka [21]

Answer:

v=10m/sec

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

F_N+Fg

where

F_N =forceof the normal

Fg= force due to gravity

Generally the net force is given to be FC(force towards center)

F_C =F_N + Fg

F_N =F_C -Fg

F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

m_g + \frac{8m}{4}= \frac{mv^2}{8}

\frac{5mg}{4} =\frac{mv^2}{8}

v^2 =\frac{40*10}{4}

v=10m/sec

Therefore the speed of the roller coaster is given ton be v=10m/sec

5 0
3 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

8 0
3 years ago
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