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3 years ago

I don’t get question number 10 for science

Physics

1 answer:

jeyben [28]3 years ago

5 0

a. matter

b. mass

c. weight

d. gravity does not change

e. determined by gravity

f. location determines amount

It's kind of hard for me to explain but to further understand this topic you can google the deference between mass and weight. Both mass and weight are both related to matter to that's why matter is a.

Weight is determined by gravity because you would way less on the moon that doesn't have that much of a gravitational pull. That's why people jump really high on the moon. that's also why the location determines the amount.

Mass however doesn't change by gravity. When you go to the moon you do not loose more of yourself or become smaller, you stay the same size. That is part of mass.

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A child looked at

fgiga [73]

The answer to your question is A.

4 0

2 years ago

How many revolutions per minute would a 25 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the

kozerog [31]

Refer to the diagram shown below.

Let ω = the angular velocity (rad/s) of the wheel.

At the topmost point, the passenger, with mass m, will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.

Therefore
mω²r = mg
ω = √(g/r)

Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s

Because 1 revolution per second is 2π rad/s, therefore
$\omega = (0.8854 \, \frac{rad}{s} )*( \frac{1}{2 \pi} \, \frac{rev}{rad} )*(60 \, \frac{s}{min} ) = 8.455 \, rev/min$

Answer: 8.455 rev/min

8 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass

vladimir2022 [97]

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

6 0

3 years ago

Can you hit the monkey at a slow speed with gravity on?

ExtremeBDS [4]

Answer:

Explanation:

we can't hit a monkey at a slow speed with gravity on

5 0

3 years ago