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lina2011 [118]
3 years ago
14

When you cut a object does the density same the same between the two objects or no and if no what will happen

Physics
1 answer:
kow [346]3 years ago
6 0
The density will always be the same.
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Over a period of operation, the useful work output of the fluorescent bulb was
Nadya [2.5K]

Answer:

199.0521 Will be the answer

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3 years ago
Where is there potential energy throughout the loading, cocking, and releasing if the trebuchet?
pshichka [43]

Answer:

adapted from NOVA, a team of historians, engineers, and trade experts recreate a medieval throwing machine called a trebuchet. To launch a projectile, a trebuchet utilizes the transfer of gravitational potential energy into kinetic energy. A massive counterweight at one end of a lever falls because of gravity, causing the other end of the lever to rise and release a projectile from a sling. As part of their design process, the engineers use models to help evaluate how well their designs will work.

Explanation:

4 0
4 years ago
An evacuated long tube contains a coin and a feather. If both objects fall together starting from the top of the tube, it is exp
Natasha_Volkova [10]

Answer:

gexp = 3.65 m/s²

Explanation:

The value of acceleration due to gravity changes with the altitude. The following formula gives the value of acceleration due to gravity at some altitude from the sea level:

gexp = g(1 - 2h/Re)

where,

gexp = expected value of g at altitude = ?

g = acceleration due to gravity at sea level = 9.8 m/s²

h = altitude = 2000 km = 2 x 10⁶ m

Re = Radius of Earth = 6.37 x 10⁶ m

Therefore,

gexp = (9.8 m/s²)(1 - 2*2 x 10⁶ m/6.37 x 10⁶ m)

<u>gexp = 3.65 m/s²</u>

5 0
3 years ago
At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2)i hat + (6.00 m/s2)j. It m
Elina [12.6K]

Answer:

r = 3519.55 m

Explanation:

We know that the acceleration of a particle in a circular motion is directed towards the center of the circumference and has magnitude:

F = rω^2

Where r is the radius of the circumference and ω is the angular velocity.

From the two acceleration vectors we find that their magnitude is

√(7^2+6^2) = √85

Therefore:

√85 m/s^2= rω^2

Now we need to calculate the angular velocity to obtain the radius. Since t2-t1 = 3s is less than one period we can be sure that the angular velocity is equals to the angle traveled between this time divided by 3 s.

The angle with respect to the x-axis  for the particle at t1 and t2 is:

\theta 1 =\cos ^{-1}\left(\frac{7}{\sqrt{85}}\right)\\\theta 2 =\cos ^{-1}\left(\frac{6}{\sqrt{85}}\right)\\

Therefore, the angular velocity ω is (in radians per second):

\omega = \frac{\theta2 - \theta1}{3 s} = 0.0511813 \frac{1}{s}

Therefore:

r = √85 / (0.0511813)^2 = 3519.55 m

4 0
3 years ago
A car travels at uniform acceleration over a tine interval of 20.0s. its initial velocity is 11.0m/s and its final velocity is 3
SIZIF [17.4K]

Since the acceleration is uniform, we can calculate it from the data we are given:

a = (vf - vi)/2

where vf=33 m/s and vi=11 m/s

Then use Suvat's equation:

x(t) = vi*t + 0.5 * a * t

where t=20s

3 0
4 years ago
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