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Alborosie
26 days ago
10

Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed

both with a piece of wool. What did Sonia observe, and why?
Physics
2 answers:
Ostrovityanka [3.2K]26 days ago
8 0
Sonia noticed that the balloons pushed away from each other. This occurs because they both gained the same type of charge after being rubbed with the wool, and similar charges repel.
Keith_Richards [3.2K]26 days ago
6 0
When Sonia rubs the balloon with a wool cloth, electrons from the balloon move to the wool because of the friction. This process causes the balloon to become positively charged. Since both balloons are treated similarly, they acquire positive charges. It’s a known principle of physics that like charges repel one another, so the two balloons will push away from each other after being rubbed with the wool.
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Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for
Ostrovityanka [3204]

2*3.5 = 7m/s

You need to multiply the acceleration by the time (which must both be in seconds; if not, convert them to the same units).

7 0
1 month ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
ValentinkaMS [3465]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is \frac{4}{3} \pi (\frac{D}{2})^3

Thus \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

The radius of the disk is 1.15(\frac{ D}{2} )

Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
1 month ago
A charge of 4 nc is placed uniformly on a square sheet of nonconducting material of side 17 cm in the yz plane. (a) what is the
Softa [3030]

The sheet's charge density calculates to be 1.384×10⁻⁷C/m².

Charge density is the ratio of charge per unit area.

The square sheet has dimensions  l=17 cm.

To compute the area A of the sheet.

A=l^2=(17 cm)^2= (17*10^-^2m)^2=0.0289 m^2

The overall charge Q on the sheet is

Q=4nC=4*10^-^9C

The charge density σ is defined as

\sigma=\frac{Q}{A}

We substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

\sigma=\frac{Q}{A}\\ =\frac{4*10^-^9C}{0.0289 m^2} \\ =1.389*10^-^7C/m^2

Thus, the charge density for the sheet is 1.384×10⁻⁷C/m².

8 0
29 days ago
You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, th
Yuliya22 [3333]

Answer:

The y-component of the position vector for the car is 670m/s.

The x-component of the acceleration vector is -3 and the y-component equals 40.

Explanation:

The car's displacement vector corresponding to the velocity

\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s

is derived from the integration of the velocity.

By integrating \boldsymbol{v}, we obtain the displacement vector \boldsymbol{d}:

\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j}  )

Assuming the vehicle's starting position is

\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})

the displacement at time t would then be

\boldsymbol{d(t)}= \boldsymbol{r+d}

\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j}  )}

At the moment t=10s, we find

\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j}  )}m

The y-component of the position vector for the car is 670m/s.

The acceleration vector can be calculated as the derivative of the velocity vector:

\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})

At t=10s, it becomes

\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2

The x-component of the acceleration vector is -3 and the y-component is 40.

5 0
2 months ago
A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s
Ostrovityanka [3204]

Explanation:

The entire system will accelerate due to the applied force. The box will feel a force opposing friction, and once this force surpasses the friction, the box will start moving. Therefore,

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The force applied is expressed as

F = (m1 + m2)×a hence

F = μs×g×(m1+m2)

3 0
1 month ago
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