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4 years ago

A convex lens can produce a real image but not a viral image a. true b. false

Physics

1 answer:

serious [3.7K]4 years ago

5 0

The answer is true a convex lens can produce a real image but not a viral image

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A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t

Grace [21]

Answer: $25.97 rad/s^2$

Explanation:

Given

radius of wheel $r=0.28 m$

mass of wheel $m=8.80 kg$

Force $F=32 N$

Moment of Inertia of solid wheel $I=\frac{mr^2}{2}$

$I=\frac{8.8\times 0.28^2}{2}$

$I=0.344 kg-m^2$

Torque is given by

$\tau =F\times r=I\times \alpha$

$32\times 0.28=0.344\times \alpha$

$\alpha =25.97 rad/s^2$

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

3 0

3 years ago

A balloon has a volume of 8.5 L at a pressure of 150 kPa. What will be the new volume when the pressure drops to 55.0 kPa?

lukranit [14]

Answer:

V₂ = 23.18 L

Explanation:

Given that,

Initial volume of a balloon, V₁ = 8.5 L

Initial pressure, P₁ = 150 kPa

We need to find new volume when the pressure drops to 55 kPa.

It is based on the concept of Boyle's law. The mathematical form of the Boyle's law is given by :

$P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{150\times 8.5}{55}\\\\=23.18\ L$

So, the new volume is 23.18 L.

6 0

3 years ago

How to solve these two questions?

hammer [34]

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$

While the area of the second square is

$A_2 = (0.002 s - 0.001 s)(0.010 A)=1\cdot 10^{-5}C$

So, the total area (and the total charge) is

$Q=A_1 +A_2 = 5\cdot 10^{-6} + 1\cdot 10^{-5} = 1.5\cdot 10^{-5}C=1.5 \mu C$

3 0

4 years ago

What’s the answer to this

ladessa [460]

Choices 1, 2, and 4 . . . . . Yes

Choices 3 and 5 . . . . . No

6 0

4 years ago

2. What does the term reflection mean?

Vesna [10]

Answer:

egrfeirugherhgourehgabgwehgoehborghrewuhgelkg

Explanation:

4 0

3 years ago