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Lady bird [3.3K]
3 years ago
7

Simon is riding a bike at 12 km/h away from his friend Keesha.He throws a ball at 5 km/h back to Keesha, who is standing still o

n a sidewalk.
 
How fast would Keesha say the ball is traveling toward her?
Physics
1 answer:
Tems11 [23]3 years ago
5 0

I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.

In fact !  You didn't even say which frame the '12 km/hr' of his
bike is measured in.

So there are several different ways this could go.  I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.

-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.

-- He throws a ball at Keesha, at 5 km/hr relative to his own face.

-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.

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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
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2 years ago
B) Smartphone, iPad, and tablet are also a kind of computer.
valentinak56 [21]

Answer:

Yes

Explanation:

7 0
2 years ago
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and han
sleet_krkn [62]

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

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3 years ago
A crowbar makes our work easier by multiplying effort true or false​
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Answer:

True

Explanation:

A crowbar makes our work easier by multiply effort because it belongs to first class lever.

And first class lever makes work easier by multiplying the effort

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A second-grade teacher dropped a box of paper clips and they scattered all over the floor. She then asked her students, "Why wil
Yuliya22 [10]

The magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips.

<h3>Attraction of magnets</h3>

The like poles of magnets repel while unlike poles of magnets attracts. Magnets attracts irons or metallic materials.

The paper clips are mettalic or made of iron and hence the magnet will attract them.

There we can conclude that the magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips and will help to gather them together for easy picking.

Learn more about magnets here: brainly.com/question/14997726

8 0
1 year ago
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