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3 years ago

Which of the following represents an image that is located behind a mirror?

Physics

2 answers:

9966 [12]3 years ago

3 0

Answer:

i think its -di........

Marina CMI [18]3 years ago

3 0

Explanation:

There are some conventions while solving the problems based on mirrors. All parameters are taken like x-y coordinate system.

All the measurements are done from the optical centre of the mirror.

$d_i$ is the image distance

$d_o$ is the object distance

$-d_i$ is the image distance which is formed behind the mirror.

$-d_o$ is the object distance when an object is placed behind the mirror.

So, $-d_i$ shows the image that is located behind the mirror. Hence, this is the required solution.

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The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration, skidding to a comp

Ber [7]

Answer: $-6.80\ ms^{-2}$

Explanation:

We know that the formula for acceleration is given by:

$a=\dfrac{v-u}{t}$ , where v = Final velocity

u= Initial velocity

Given : The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration.

i.e. u= 110 km/h $=\dfrac{110\times1000}{3600}\approx35.6\ m/s$ [∵ 1 km= 100 meters and 1 hour = 3600 seconds]

v= 0 m/s ( At brake , final velocity becomes 0)

t=4.5 seconds

Substitute all the values in the formula , we get

$a=\dfrac{0-30.56}{4.5}\approx-6.80\ ms^{-2}$

Hence, the average acceleration of the car during braking is $-6.80\ ms^{-2}$ .

5 0

2 years ago

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 7.0 s

marta [7]

Answer

given,

initial speed of merry-go-round = 0 rad/s

final speed of merry-go-round = 1.5 rad/s

time = 7 s

Radius of the disk = 6 m

Mass of the merry-go-round = 25000 Kg

Moment of inertia of the disk

$I = \dfrac{1}{2}MR^2$

$I = \dfrac{1}{2}\times 25000\times 6^2$

I = 450000 kg.m²

angular acceleration

$\alpha = \dfrac{\omega_f-\omega_0}{t}$

$\alpha = \dfrac{1.5-0}{7}$

$\alpha =0.214\ rad/s^2$

we know,

$\tau= I \alpha$

$\tau= 450000\times 0.214$

$\tau=96300\ N.m$

8 0

3 years ago

The model of the atom proposed by Greek philosophers appears similar to the model proposed centuries later by Dalton. What was t

Sladkaya [172]

The main difference between the model of the atom proposed by Greek philosophers and the model proposed centuries later by Dalton is that the Greek one was mainly speculative and philosophical - it wasn't based on real evidence, but on their suggestions and thoughts about the matter. On the other hand, Dalton had the means to prove his theory using viable evidence, not just speculations.

4 0

3 years ago

A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a

chubhunter [2.5K]

Answer:

The ballon will brust at

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Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

$R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}$

Rmax = 10.001 cm

Therefore, the max volume will be:

$V_{max} = \frac{4}{3} \pi R_{max}^3$

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

$P_{max} = P_1 \frac{V_1T_2}{V_2T_1}$

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0

2 years ago

If an object travels 15 metres in 5 seconds, what is the slope of its distance-time graph?

SOVA2 [1]

Answer:

3 m/s

Explanation:

The slope is distance divided by time, also known as 'speed'.

15 meters / 5 seconds = 3 meters per second.

8 0

2 years ago