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zlopas [31]
2 years ago
11

What would be the position of the centre of mass?

Physics
1 answer:
AnnyKZ [126]2 years ago
6 0

Answer: the average position of all the parts of the system, weighted according to their masses.

Explanation:

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Mrs. Rohaley gives you a piece of iron and asks you to determine the physical and chemical properties of the metal. She wants yo
Fofino [41]

Answer:The answer is B

Explanation:

5 0
3 years ago
What is championship game of baseball called? The World Series The Super Bowl The World Cup​
emmasim [6.3K]

Answer:

The World Series

Explanation:

The Super Bowl is the championship American Football game, and the World Cup is the Soccer/Football game.

American Football and Football are different things. The first is what Americans call football, while the other is what Americans call soccer. It is confusing.

6 0
3 years ago
A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu
Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

                                                     = 25° C

The specific heat capacity of aluminium, c = 900  J/kg°C

The formula for thermal energy,

                             <em>Q = mcΔT</em>

                                 = 1.0 x 900 x 25

                                 = 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0
3 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\&#10;x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

5 0
3 years ago
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