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3 years ago

I traveled 6.00 km to school in 0.250 hours . What was my average speed

Physics

1 answer:

SOVA2 [1]3 years ago

5 0

Answer: 6.67 m/s

Explanation:

Recall that average speed is the total distance covered by a moving body divided by the total time taken. The SI unit is metres per second.

i.e Average speed = distance/time taken

Given that:

Distance covered = 6.00 km

Convert kilometers to meters

If 1 km = 1000 m

6.00 km = (6.00 x 1000) = 6000m

Time taken to school = 0.250 hours

Convert time in hours to seconds

If 1 hour = 60 minutes & 1 minute = 60 seconds

Then, 0.250 hours = (0.250 x 60 x 60)

= 900 seconds

Average speed = ?

Average speed = 6000m/900s

= 6.67 m/s

Thus, your average speed to school is 6.67 m/s

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You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s

BlackZzzverrR [31]

1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>

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3 years ago

A violin string vibrates at 260 Hz when unfingered. At what frequency will it vibrate if it is fingered one fourth of the way do

Advocard [28]

Answer:

346.66 Hz

Explanation:

$l_1$ = Length of string which is unfingered = l

$l_2$ = Length of string which is vibrate when fingered = $l-\dfrac{1}{4}l=\dfrac{3}{4}l$

$f_1$ = Unfingered frequency = 260 Hz

$f_2$ = Fingered frequency

Frequency is inversely proportional to length

$f=\dfrac{1}{l}$

So,

$\dfrac{f_1}{f_2}=\dfrac{l_2}{l_1}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{\dfrac{3}{4}l}{l}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{3}{4}\\\Rightarrow f_2=\dfrac{4}{3}f_1\\\Rightarrow f_2=\dfrac{4}{3}260\\\Rightarrow f_2=346.66\ Hz$

The frequency of the fingered string is 346.66 Hz

3 0

2 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

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Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to

Degger [83]

The acceleration of the box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

m is the mass of the box
a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

θ is the inclination of the plane
μ is coefficient of friction

F(net) = 170 - (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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10 months ago

vodka [1.7K]

Answer:

3 electron hai bro of puch mujhe sab aata h

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2 years ago

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