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dexar [7]
3 years ago
6

Two soccer players, mary and jane, begin running from approximately the same point at the same time. mary runs in an easterly di

rection at 3.59 m/s, while jane takes off in a direction 42.8 ◦ north of east at 4.67 m/s. what is the magnitude of the velocity of jane relative to mary?

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0
As you illustrate the problem, you may come up with the same diagram as the one shown in the attached picture. The relative velocity is the linear velocity between the two velocities. This is calculated by simply finding the difference of their velocities. Since, the velocity is with respect to Jane relative to Mary, the solution is as follows:

Relative velocity = Velocity of Jane - Velocity of Mary
Relative Velocity = 4.67 m/s - 3.59 m/s = 1.08 m/s 

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Radio telescopes beam radio waves into space. To calculate the speed of the radio waves in space, which information, if any, is
Anastaziya [24]

Answer:

No more information is needed

Explanation:

Radio waves are electromagnetic energy, lower frequency forms of this type of energy that includes light and cosmic rays on the high frequency end that we are able to detect. So in free space (vacuum), radio waves travel at their fastest velocity, the “speed of light”. The reason for the quotation marks is because when light or radio waves are propagating through matter, we observe them traveling more slowly.

7 0
3 years ago
Friction force affects the work done by the applied force.<br><br>T OR F ​
lakkis [162]

Answer:

TRUE.

Explanation:

YES IT EFFECTS.

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7 0
3 years ago
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A5 cm object is 18.0 cm from a convex lens, which has a focal length of 10.0 cm.
Masja [62]

Explanation:

We have,

Height of object is 5 cm

Object distance from a convex lens is 18 cm

Focal length of convex lens is 10 cm

i.e. h = 5 cm

u = -18 cm

f = +10 cm

Let v is distance of the image from the lens. Using lens formula :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{(-18)}\\\\v=22.5\ cm

The magnification of lens is :

m=\dfrac{v}{u}=\dfrac{h'}{h}, h' is height of the image

\dfrac{v}{u}=\dfrac{h'}{h}\\\\h'=\dfrac{vh}{u}\\\\h'=\dfrac{22.5\times 4}{(-18)}\\\\h'=-5\ cm

h' = -5.00 cm (in three significant figures)

6 0
3 years ago
A 1500-W heater is connected to a 120-V line for 2.0 hours. How much heat energy is produced?
weeeeeb [17]
<span>A joule is equal to one watt per second and so we must find out how many seconds are in two hours. A shortcut you can use is remove the zeros and multiplying the remaining numbers before adding the the zeros back to the total of your result 6 x 6 x 2 = 72 and thus 60 x 60 x 2 = 7200 72 x 15 = 1080 and thus 7200 x 1500 = 10800000 joules one mega joule equals one million joules So we can simplify things and say 10.8 MJ</span>
3 0
3 years ago
An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0
Mazyrski [523]

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

\theta=0.260t^2

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t

Then, you evaluate for t=2.40 s:

\omega=0.52(2.40)=1.248\frac{rad}{s}

(b) The linear velocity is calculated by using the following formula:

v=\omega r

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}

(c) The tangential acceleration is:

a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}

(d) The radial acceleration is:

a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}

4 0
3 years ago
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