Ask question

Ask question

All categories

2 years ago

5

camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin

g this information and the photo itself, approximately how fast did the car drive

1 answer:

son4ous [18]2 years ago

7 0

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

$v=\frac{\Delta x}{\Delta t}$

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

$v=\frac{37.1}{5.4}$

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

You might be interested in

Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show

Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call $R_M$ its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

$R_M = 1000 R$ (1)

Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

$V_M = V_R$

Using Ohm's law, $V=RI$ , we can rewrite the previous equation as:

$R_M I_M = R I_R$

where

$I_M$ is the current in the meter

$I_R$ is the current in the bulb

Using (1), this equation becomes

$(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}$

so, the current in the meter is 1000 times less than through the bulb.

5 0

2 years ago

Which title best reflects the main idea of the passage? The Role of Convection in the Distribution of Earth's Energy The Role of

Leto [7]

Answer:

The Role of Heat Transfer Methods in the Distribution of Earth's Energy

Explanation:

8 0

2 years ago

Read 2 more answers

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0

2 years ago

Bianca is standing at x =600m. Firecracker 1, at the origin, and firecracker 2, at x =900m, explode simultaneously. The flash fr

agasfer [191]

Answer: $3 \mu s$

Explanation:

Given

Bianca is at $x=600 m$

i.e. distance between origin and Bianca is $600 m$

time taken to reach Bianca eyes is

$t=\frac{600}{speed\ of\ light}$

$t=\frac{600}{3\times 10^8}$

$t=2\times 10^{-6} s$

$t=2 \mu s$

i.e. Cracker exploded at $t=2\mu s$ because it is observed at $t=4\mu s$

Time taken by second cracker flash to reach Bianca eyes

$t_2=\frac{300}{3\times 10^8}$

$t_2=10^{-6}$

$t_2=1 \mu s$

Therefore it will be observed at $t=3 \mu s$

4 0

2 years ago

A cup slides off a 1.1\,\text m1.1m1, point, 1, start text, m, end text high table with a speed of 1.3\,\dfrac{\text m}{\text s}

Tju [1.3M]

Answer:

<h3>0.145m</h3>

Explanation:

Using the equation of motion formula y = ut + 1/2gt² where;

y is the horizontal displacement

u is the initial velocity

g is the acceleration due to gravity

t is the time

To calculate the horizontal displacement, we need to first get the time t. Using the equation:

v = u+gt

1.3² = 0+(9.8)t

1.69 = 9.8t

t = 1.69/9.8

t = 0.172s

Substituting the time into the equation above to get y

y = 0+1/2(9.8)(0.172)²

y = 4.905(0.029584)

y = 0.145m

Hence the cup's horizontal displacement during the fall is 0.145m

8 0

2 years ago