Please help me with my question

Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber

puteri [66]

Answer:

C. A rubber rod and a glass rod charged this way have opposite charges on them.

Explanation:

When a rubber rod is rubbed against cat fur, it acquires a negative charge, it becomes negatively charged.

When you then try to bring two rubber rod's together, they repel because like charges repel.

Meanwhile, when you rub a glass rod against silk, it loses electrons to the silk material and becomes positively charged.

When you bring two positively charged glass rod's together, they repel, because like charges repel.

However, when you bring the rubber rod and a glass rod together, the attract each other because unlike/opposite charges attract.

5 0

3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:

$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

 $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

6 0

3 years ago

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s

Gekata [30.6K]

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

$P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580$

7 0

3 years ago

A radio-controlled car increases its kinetic energy from 4 j to 12 j over a distance of 2 m. what was the average net force on t

just olya [345]

The answer to this is easy once you look at the units for Joules. 1 Joule = 1 N.m (Newton.meter). The 'Newton' is the units of force that we are trying to find, and we know the meters is 2, from the question. So you have an 8Joule or 8N.m energy difference over 2 meters.

well if we know the meters, then the real question is written as:
8N.m = ?N x 2m
so just solve for N;
N = 8N.m / 2m = 4
So F = 4N

3 0

4 years ago

A cube of iron (Cp = 0.450 J/g•°C) with a mass of 55.8 g is heated from 25.0°C to 49.0°C. How much heat is required for this pro

Rus_ich [418]

Q = mcΔT
q = 55.8g x 0.450J/gC x 23.5C 
q = 590. J ................ to three significant digits

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

8 0

3 years ago

Read 2 more answers