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3 years ago

Can anyone help??????????????

Physics

1 answer:

DerKrebs [107]3 years ago

3 0

I think the answer should be the last one. Magnets attract magnets with unlike poles and repel magnets with like poles

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(50 POINTS) please answer each question:

Advocard [28]

Answer:

i know the first one.

Explanation:

the distance from earth in light years is 6 trillion miles.

6 0

2 years ago

Select the appropriate shape for the given volume form

vredina [299]

Fist one is a cylinder
the second, i believe is a sphere
the third is a rectangular prism
and the last is the same as the first, a cylinder

3 0

2 years ago

How can a scientist ensure that his or her data are reliable?

laiz [17]

If the scientist repeats the experiment over and over and gets the same results. Also if the scientist peer reviews the experiment to make sure there is no bias in his or her results.

7 0

3 years ago

Read 2 more answers

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

Oksana_A [137]

Answer:

a) $\omega=1.36rad/s$

b) $\omega=12.99rpm$

c) $F=705.6N$

Explanation:

a) The angular velocity is related to the centripetal acceleration by the formula $a_{cp}=\omega^2r$ , which for our purposes we will write as:

$\omega=\sqrt{\frac{a_{cp}}{r}}$

Since <em>we want this acceleration to be 1.5 times that due to gravity</em>, for our values we will have:

$\omega=\sqrt{\frac{1.5g}{r}}=\sqrt{\frac{(1.5)(9.8m/s^2)}{(8m)}}=1.36rad/s$

b) 1 rpm (revolution per minute) is equivalent to an angle of $2\pi$ radians in 60 seconds:

$1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s$

Which means <em>we can use the conversion factor</em>:

$\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1$

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

$\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm$

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

$F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N$

8 0

3 years ago

A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

Lilit [14]

Answer:

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut + $\frac{1}{2}$ a $_y$ t²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850 + $\frac{1}{2}$ ×a $_y$ ×(0.850)²

1.29 = 0.36125a $_y$

a $_y$ = 1.29 / 0.36125

a $_y$ = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a $_y$

where m₁ is the mass of the text book ( 2.10 kg )

a $_y$ is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N

3 0

2 years ago