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3 years ago

_____ lenses always produce smaller images

Physics

1 answer:

tatyana61 [14]3 years ago

4 0

Answer:

Explanation:

The Answer is Concave. Concave images make images smaller, they refract the light rays in many directions, causing the original image to shrink. This is why shortsighted people wear concave lenses. Convex lenses, the other type of diverging lenses, cause the image to be larger, because it concentrates the image in a specific point. That is how a magnifying glass works.

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What causes resistance in an electric circuit? How is resistance measured?

Sergeeva-Olga [200]

Resistance can be caused by specific resistance tools, like resistors, or loads (the part of the component that uses the electricity) like a light bulb or a speaker. Resistance is measured in ohms and can be measured by a multimeter.

6 0

3 years ago

Mary spots Bill approaching the dorm at a constant rate of 2.00 m/s on the walkway that passes directly beneath Mary's window, 1

vagabundo [1.1K]

Answer:

Mary will have to wait for 63.2 seconds

Explanation:

Time required for the apple to drop from a height of 17.0 m above the ground to 1.65 m above the ground is given by the formula below:

t = √2h/g where h is height through which the object falls, g is acceleration due to gravity

h = 17.0 - 1.65 = 15.35 m

g = 9.8 m/s²

t = √(2 * 15.35/9.8)

t = 1.77 s or approximately 1.8 s

Time taken for bill to get to the point below Mary's window is given below;

time taken = distance/velocity

distance = 130 m; velocity = 2.0 m/s

time taken by Bill = 130/2.0 = 65 s

Therefore, Mary will have to wait for (65 - 1.8) s = 63.2 seconds

4 0

4 years ago

A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another

laiz [17]

Answer

given,

watermelon blown into three pieces

two pieces of mass m

both pieces speed = v = 31 m/s

mass of third piece = 3 m

using conservation of mass

$0 = m V \hat{i} + m V \hat{j} + (3\ m)V_3$

$V_3 = \dfrac{V}{3}(-\hat{i}-\hat{j})$

$V_3 = \dfrac{V}{3}\sqrt{1^2+1^2}$

$V_3 = \dfrac{V}{3}\sqrt{2}$

velocity of third component

$V_3 = \dfrac{31}{3}\sqrt{2}$

$V_3 = 14.61\ m/s$

angle

$\theta = tan^{-1}(\dfrac{-1}{-1})$

$\theta = tan^{-1}(1)$

$\theta = 45^0$

5 0

4 years ago

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm

nlexa [21]

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $\frac{4}{3} \pi r^{3}$

volume of balloon = $\frac{4}{3}$ x 3.142 x $24^{3}$ = 57913.34 m^3

using the gas equation,

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$\frac{0.03*57913.34}{200}$ = $\frac{1*V2}{349}$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3}$ x 3.142 x $r^{3}$ = 3031.76

4.19 $r^{3}$ = 3031.76

$r^{3}$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

4 0

3 years ago

If v =5.00 meters/second and makes an angle of 60 with the negative direction of the y-axis coalculate the possible values of vx

Sever21 [200]

An angle of 60 degrees with the negative y-axis could mean 60 degrees clockwise or counterclockwise, which translates to two possible angles (starting from the positive x-axis and moving counterclockwise) of 210 degrees or 330 degrees.

Then the horizontal component $v_x$ of a velocity vector $\mathbf v$ with magnitude $5.00\,\dfrac{\mathrm m}{\mathrm s}$ could be one of two expressions:

$v_x=\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\cos210^\circ=-4.33\,\dfrac{\mathrm m}{\mathrm s}$

$v_x=\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\cos330^\circ=4.33\,\dfrac{\mathrm m}{\mathrm s}$

3 0

3 years ago