Ask question

Ask question

All categories

sertanlavr [38]

2 years ago

13

Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by

the unaided eye. c) A virtual image cannot be photographed. d) A real image must be erect. e) Mirrors always produce real images because they reflect light.

1 answer:

ankoles [38]2 years ago

5 0

Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

A virtual image can not be formed on a screen.

For image:

1.A virtual image can be viewed by the unaided eye.

2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

So the option A is correct.

You might be interested in

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

horrorfan [7]

Answer:

The value is $t_1 = 9 \ s$

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

$v = u + at$

Here u = 0 \ m/s

a = 13.5 $m/s^2$

So

$v = 0 + 13.5 * 3.10$

=> $v = 41.85 \ m/s$

The is distance it covers at this time is

$s = u * t + \frac{1}{2} a * t^2$

=> $s = + \frac{1}{2} * 13.5 * 3.10^2$

=> $s =64.87$

Now when sled stops its the final velocity is $v_f = 0 m/s$ while the initial velocity will be the velocity after its acceleration i.e $v = 41.85 \ m/s$

So

$v_f = v + a_1t_1$

Here $a_1 = - 4.65$ , the negative sign shows that it is deceleration

So

$0 = 41.85 - 4.65 * t_1$

=> $t_1 = 9 \ s$

3 0

2 years ago

Find the mass of a person walking west at a speed of 0.8 m/s with a momentum of 52.0 kg.m/s west.

KIM [24]

Answer:

mass of the person walking to west is 65 kg.

Given:

Momentum = 52 $\frac{kg m}{s}$

Speed = 0.8 $\frac{m}{s}$

To find:

Mass of the person = ?

Formula used:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Solution:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Mass = $\frac{P}{v}$

m = $\frac{52}{0.8}$

m = 65 kg

Thus, mass of the person walking to west is 65 kg.

5 0

1 year ago

A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is

Step2247 [10]

Answer:

The decelerating force is $3\times 10^{- 11}\ N$

Solution:

As per the question:

Frontal Area, A = $10\ m^{2}$

Speed of the spaceship, v = $1\times 10^{6}\ m/s$

Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

A = Area

v = velocity

t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

4 0

1 year ago

Researchers interested in studying stress gave 150 high school seniors a very difficult math exam. After the test, the researche

belka [17]

Answer:

<u>Informed Consent- The study didn't conform this ethical standard.</u>

<u>Debriefing- The study conforms this ethical standard.</u>

<u>Confidentiality- The study conforms this ethical standard.</u>

<u>Protection from harm- The study conforms this ethical standard.</u>

Explanation:

<u>Informed consent</u> is described as a procedure whereby researchers tend to provide details about specific research they are going to conduct, the risk & benefits involved in the study, different alternatives involved in the procedure, etc.

<u>Debriefing </u>is described as a process that is being conducted in any of the different psychological research encompassing human participants after specific research is completed. The researcher tends to describe the details of the research to the participants.

<u>Confidentiality</u> is described as one of the different code of ethics followed by health workers or psychologists. While practicing confidentiality, a psychologist tends to promise his or her participants that he or she will keep everything a secret whatever is being discussed or shared between both of them.

<u>Protection from harm</u> determines that the psychologists conducting research follows the ethics in which he or she has to protect all the participants from any kind of harm.

4 0

1 year ago