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3 years ago

6

How do you find the volume of an irregular shaped object

1 answer:

mash [69]3 years ago

8 0

You can find the volume of an irregular shaped object by immersing it in water in a beaker or other container with volume markings, and by seeing how much the level goes up.

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Which of these experiments would make use of qualitative data

7nadin3 [17]

It has to due with numbers so I would say the last one!

5 0

3 years ago

Read 2 more answers

An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution

givi [52]

With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
= (15π rad/s)*(1 rev/2π rad)
= 7.5 rev/s

Answer: 7.5 revolutions per second.

7 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

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Aneli [31]

$d^2=4^2+3^2\\\\d^2=16+9\\\\d^2=25\\\\d=5$

3 0

3 years ago

A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F

vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

$t = \frac{d}{v_{net}}$

here net speed of the plane will be given as

$v_{net} = v + v_w$

$t = \frac{1000}{205 + v_w}$

similarly when it moves under the condition of headwind its net speed is given as

$v_{net} = v - v_w$

now time taken to cover the distance is 2 hours more

$t + 2 = \frac{1000}{205 - v_w}$

now solving two equations

$\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}$

solving above for v_w we got

$v_w = 40.4 mph$

6 0

3 years ago