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Bad White [126]
3 years ago
15

An object has a velocity of 10m/s south. in what direction is its momentum

Physics
1 answer:
Tanya [424]3 years ago
7 0
The direction is from the west
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A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured
nikklg [1K]

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

4 0
3 years ago
Jenny and Betty are having a great time at Busch Gardens riding the Ubanga Banga bumper cars. Jenny, who is traveling southward
Alex777 [14]

Jenny is traveling southward.  In order to stop, she needs a northward acceleration.


A better way to say it:

Jenny is traveling southward in her bumper car, so the direction of her velocity is south.  In order to reduce her velocity to zero, a velocity of equal magnitude but directed north must be added to it.  Then the change in velocity is positive northward, and the change in velocity per unit time is acceleration.

6 0
3 years ago
Read 2 more answers
Please help on this one?
Sati [7]
I’m pretty sure it’s A
4 0
4 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16
ycow [4]
We assign the variables: T as tension  and x the angle of the string
 The  <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx. 
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
4 0
3 years ago
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An electron in a mercury atom changes from energy level b to a higher energy level when the
emmainna [20.7K]
The energy is 3.06 electronvolts,  E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J


4 0
3 years ago
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