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erma4kov [3.2K]
2 years ago
14

an object is held 33.5 cm from a convex mirror. it creates a virtual image of magnification 0.253. what is the focal length of t

he mirror?
Physics
1 answer:
MaRussiya [10]2 years ago
7 0

Answer:

Magnification= -image distance/object distance

.253=image distance/33.5

image distance= 8.48 cm

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Elena-2011 [213]


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8 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
PLEASE ANSWER QUICKLY
Rom4ik [11]

Answer:

a luminous ball of plasma

6 0
3 years ago
Pls help mee
NISA [10]

Answer:

the Arrow X shows the direction of amplitude

Explanation:

As the amplitude is the maximum displacement of a wave from the mean position

8 0
2 years ago
A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th
lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

7 0
3 years ago
Read 2 more answers
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