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2 years ago

Can electrostatic attract both pull and pull?

Physics

2 answers:

Delvig [45]2 years ago

8 0

Answer: Yes

Explanation:

Electrostatic forces are non-contact forces; they pull or push on objects without touching them. Rubbing some materials together can result in something called 'charge' being moved from one surface to the other. Charged objects pull on other uncharged objects and may either push or pull on other charged objects

Sedaia [141]2 years ago

6 0

Yes they can attract both pulls

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A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it

levacccp [35]

Answer:

$46620\ \text{N}$

Explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Normal force at the point will be

$N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}$

The force exerted on the track is $46620\ \text{N}$ .

8 0

3 years ago

The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are l

marshall27 [118]

Answer:

C)the right cord pulls on the pulley with greater force than the left cord

Explanation:

As we can see the figure that B is connected to the right string while A is connected to the left string

Now force equation for B as it will move downwards is given as

$(2m)g - T_B = (2m)a$

similarly for block A which will move upwards we can write the equation as

$T_A - mg = ma$

now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks

so here pulley will rotate clockwise direction

So tension in the right string must be more than the left string

So correct answer will be

C)the right cord pulls on the pulley with greater force than the left cord

3 0

2 years ago

16)

VMariaS [17]

Answer:

a.proton, proton

hope this helped :) have a goodday

6 0

3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago

Which of the following best describes the circuit shown below?

Rudiy27

Answer:

Short circuit

Explanation:

The given figure shows a short circuit. It is defined as the circuit which allows the flow of electric current when there is no resistance. It shows a battery, bulb and connecting wires.

The wire across the bulb is connected from one terminal to another without any resistance in between them.

So, the correct option is (d) " short circuit ".

3 0

3 years ago

Read 2 more answers