All categories

3 years ago

Describe 3 physical properties of this object (color, state of matter, shape, size, hardness, etc)

Physics

2 answers:

ivanzaharov [21]3 years ago

6 0

Answer: The color is orange, the state of matter is liquid

Explanation:

mezya [45]3 years ago

4 0

The color is orange, the shape is a cylinder, it is a liquid

You might be interested in

Shareen finds that when she drives her motorboat upstream she can travelwith a speed of only 8 m/s, while she moves with a speed

ANEK [815]

Let vb be the velocity of the motorboat and let vs be the velocity of the stream.

We know that when she drives upstream the velocity is 8 m/s, in this scenario the velocities point in opposite directions, then we have the equations:

$v_b-v_s=8$

When she drives downstream the velocites point in the same direction then we have the equation:

$v_b+v_s=12$

hence we have the system of equations:

$\begin{gathered} v_b-v_s=8 \\ v_b+v_s=12 \end{gathered}$

Solving the first equation for the velocity of the boat we have:

$v_b=8+v_s$

Plugging this in the second equation we have:

$\begin{gathered} 8+v_s+v_s=12 \\ 2v_s=4 \\ v_s=\frac{4}{2} \\ v_s=2 \end{gathered}$

Therefore, the velocity of the stream is 2 m/s

5 0

1 year ago

Rank the following in order of increased size? A. Proton B. Nucleus C. Electron D. Atom

Basile [38]

From largest to smallest- atom, nucleus, proton and electron

7 0

3 years ago

A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the

antoniya [11.8K]

Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

6 0

3 years ago

A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen

saw5 [17]

Answer:

The electric field strength is $4.5\times 10^{4} N/C$

Solution:

As per the question:

Area of the electrode, $A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}$

Charge, q = 50 nC = $50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m$

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

$\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}$

Now, the electric field strength of the electrode is:

$\vec{E} = \frac{\sigma}{2\epsilon_{o}}$

where

$\epsilon_{o} = 8.85\times 10^{- 12} F/m$

$\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}$

$\vec{E} = 4.5\times 10^{4} N/C$

7 0

3 years ago

A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is droppe

pshichka [43]

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

$E_p = mgh$

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

$E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J$