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Irina-Kira [14]

3 years ago

7

. An object 8.5 cm high is placed 28 cm from a converging lens. The focal length of the lens is 12 cm. Calculate the image dista

nce, di. Calculate the image height, hi.

1 answer:

velikii [3]3 years ago

5 0

Answer:

50 do

Explanation:

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From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ

V125BC [204]

Answer:

55.8 m/s

Explanation:

$f$ = Actual frequency of sound emitted by car

$f'$ = frequency observed when the car moves = $(0.86)f$

$V$ = Speed of sound = 343 m/s

$v$ = Speed of car

Frequency observed is given as

$f' = \frac{Vf}{V+v}$

$(0.86) = \frac{(343)}{343 + v}\\(0.86) = \frac{(343)}{343 + v}\\294.98 + (0.86) v = 343 \\v = 55.8 ms^{-1}$

6 0

3 years ago

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

$s=ut+\dfrac{1}{2}at_{A}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

$t_{A}=\sqrt{\dfrac{30\times2}{1.6}}$

$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

4 0

4 years ago

How does an atom change if all of its electrons are removed?

Alex

<span>The atom becomes positively charged.

When you add electrons to a neutral atom, it is no longer a neutral atom, it has a negative change and is an anion. When you take away electrons from a neutral atom, it is no longer a neutral atom- it becomes a positive atom, and is a cation.</span>

6 0

3 years ago

a) Calculate the magnitude of displacement of the car in 40 seconds. b) During which part of the journey was the car acceleratin

Zielflug [23.3K]

Answer:

a) 600 meters

b) between 0 and 10 seconds, and between 30 and 40 seconds.

c) the average of the magnitude of the velocity function is 15 m/s

Explanation:

a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:

$Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m$

b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.

Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.

c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

$Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx$

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

$Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}$

7 0

3 years ago

If you pull horizontally on a desk with a force of 150 N and the desk doesn't move, the friction force must be 150 N. Now if you

s344n2d4d5 [400]

Answer:

The friction force is 250 N

Explanation:

The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):

$\sum F=ma$

In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:

$\sum F = 0$ (1)

We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:

- the pull, forward, F = 250 N

- the friction force, backward, $F_f$

Given (1), we have

$F-F_f = 0$

So the force of friction must be equal to the pull:

$F=F_f = 250 N$

8 0

3 years ago