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3 years ago

What is the avarage velocity of a car that travels 30 kilometers due west in 0.5 hours

2 answers:

6 0

Velocity is distance over time $v= \frac{d}{t}$ . If it wants it in km/h you're good to go otherwise if its m/s you need to convert km to m by multiplying 30 by 1000 and t in hours to seconds by multiplying 0.5 by 3600.

Nikolay [14]3 years ago

5 0

Average velocity=distance/time
average v=s/t
averagev=0.3/30 min

so use your calculator 30min into seconds(30/60)
and divide your kilometre with your time(v=distance/time

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Valentin [98]

Explanation:

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1 year ago

Weigt A box sitting still on the ground by itself has a weight of 700N, what is the mass? (gravity's acceleration is 9.80 m/s2)

Alex

Force=700N
Acceleration=9.8m/s^2

Using newtons law

$\\ \rm\longmapsto F=ma$

$\\ \rm\longmapsto m=\dfrac{F}{a}$

$\\ \rm\longmapsto m=\dfrac{700}{9.8}$

$\\ \rm\longmapsto m=71.4kg$

7 0

2 years ago

An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-dia

11111nata11111 [884]

Answer:

1.4E-3J

Explanation:

Given that

Time = 8hrs = 28.8E3 seconds

Intensity= 90dB

D= 0.008m

Radius= 0.004m

So intensity is sound level Bis

10dBlog(I/Io)

I= 10 (B/10dB)Io

= 10( 90/10) x 10^-12

=0.001W/m²

But we know that

I = P/A

P= I πr²

= 5.02 x10^-8W

But energy is power x time

So E= 5.02E-8 x 28.8E3

= 1.4E-3J

5 0

2 years ago

Read 2 more answers

The dimension line is connected to the part being measured by _______________. A. Hidden lines B. Extension lines C. Visible lin

kondaur [170]

B: Extension Lines! You could have just searched this up on google

3 0

3 years ago

Read 2 more answers

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

2 years ago