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Genrish500 [490]

3 years ago

7

An object that’s charged has more electrons than protons. An object that’s charged has fewer electrons than protons. An object t

hat’s charged has the same number of electrons and protons.

1 answer:

Levart [38]3 years ago

3 0

An element is neutral when the charged is equal to zero. This means that the number of electrons and protons are equal. When an object has a NEGATIVE charge, its electrons are more than its protons. Also, the object has a POSITIVE charge when the number of its electrons are fewer compared to the number of the protons.

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Incident energy is defined as the amount of thermal energy impressed on a surface, at a certain distance from the source, genera

nignag [31]

Answer: One of the units used to measure incident energy is calories per centimeter squared (cal/cm2).

Explanation: Incident energy this is defined as the amount of thermal energy impressed on a surface, at a certain distance from the source, generated during an electrical arc event.

The working distance is the distance from where the worker stands to the source location. The most common distance for which incident energy has been determined in tests is 18 inches.

3 0

3 years ago

What type of cloud is commonly seen after a hurricane? explain why

andreev551 [17]

Cumulonimbus clouds are commonly seen after a hurricane. Hope I helped!

7 0

2 years ago

A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a

PIT_PIT [208]

Answer: $\frac{5L}{6}$

Explanation:

Given

Wooden board is pivoted at center and

Older child of mass $M=32\ kg$ is sitting at a distance of L from center

if two child of mass $\frac{M}{2}$ is sitting at a distance $\frac{L}{6}$ and $x$ (say) from pivot then net torque about pivot is zero

i.e.

$\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx$

as $\tau_{net}=0$

Therefore

$MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx$

$L-\frac{L}{6}=x$

$x=\frac{5L}{6}$

Therefore another child is sitting at a distance of $\frac{5L}{6}$

3 0

2 years ago

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

olganol [36]

Answer:

The force is $F_c = 789.03 \ N$

Explanation:

From the question we are told that

The tangential resistive force is $F_t = 115 \ N$

The mass of the wheel is m = 1.80 kg

The diameter of the wheel is $d = 50.0 cm = 0.5 \ m$

The diameter of the sprocket is $d_c = 8.50 \ cm =0.085 \ m$

The angular acceleration considered is $\alpha = 4.30\ rad/s^2$

Generally the radius of the wheel is

$r = \frac{d}{2}$

=> $r = \frac{0.5}{2}$

=> $r = 0.25 \ m$

Generally the radius of the sprocket is

$r_c = \frac{d_c}{2}$

=> $r_c = \frac{0.085}{2}$

=> $r_c = 0.0425 \ m$

Generally the moment of inertia of the wheel is mathematically represented as

$I = m * r^2$

=> $I = 1.80 * 0.25^2$

=> $I = 1.1125 \ kg \cdot m^2$

Generally the torque experienced by the wheel due to the forces acting on it is mathematically represented as

$\tau = F_c * r_c - F_t * r$

Here $F_c$ is the force acting on the sprocket

So

$\tau = F_c * 0.0425 - 115 * 0.25$

$\tau = 0.0425F_c - 28.75$

Generally the torques that will cause the wheel to move with $\alpha = 4.30\ rad/s^2$ is mathematically represented as

$\tau = I * \alpha$

So

$0.0425F_c - 28.75 = I * \alpha$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$F_c = 789.03 \ N$

5 0

3 years ago

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current

Komok [63]

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$

or

$I=\frac{V}{R}$ (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as:

$A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}$

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

$R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega$

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

$I=\frac{0.7}{14.96056465}\approx0.047A$

7 0

3 years ago