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3 years ago

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Which type of neuron stimulates muscles to contract?

1 answer:

lakkis [162]3 years ago

4 0

Answer:

a motor neuron

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An interplanetary spacecraft is moving at

mixer [17]

The awnser is. 1728000 kilometers

7 0

3 years ago

Read 2 more answers

A 500 kg object is hanging from a spring attached to the ceiling. If the spring constant in the spring is 900 N/kg, how far does

My name is Ann [436]

We are given that a 500 kg object is hanging from a spring. To determine the amount the spring is stretched we will use Hook's law, which states the following:

$F=kx$

Where:

$\begin{gathered} F=\text{ force} \\ k=\text{ spring constant} \\ x=\text{ distance stretched} \end{gathered}$

Since the object is hanging the only force acting on the spring is the weight of the object. The weight of the object is:

$F_g=mg$

Where:

$\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}$

Plugging in the values we get:

$F_g=(500\operatorname{kg})(9.8\frac{m}{s^2})$

Solving the operations:

$F_g=4900N$

Now we solve for "x" from Hook's law by dividing both sides by "k":

$\frac{F}{k}=x$

Now we plug in the known values:

$\frac{4900N}{900\frac{N}{m}}=x$

Solving the operations:

$5.4m=x$

Therefore, the spring is stretched by 5.4 meters.

7 0

1 year ago

Light of wavelength 500nm shines on a pair of slits. This produces an interference pattern on a screen 2.00m behind the slits. M

grandymaker [24]

Answer:

The value is $d = 0.000125 \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9 } \ m$

The distance is of the screen is $D = 2.0 \ m$

The width of the fifth dark spot is $y = 4.00 \ cm = 0.04 \ m$

Generally the width of a fringe is mathematically represented as

$y = \frac{ n * \lambda * D }{d }$

=> $0.04 = \frac{ 5 * 500 *10^{-9} * 2 }{d}$

=> $d = \frac{ 5 * 500 *10^{-9} * 2 }{0.04}$

=> $d = 0.000125 \ m$

7 0

3 years ago

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea

Semmy [17]

Answer:

${\rm K} = 2.4\times 10^{-19}~J$

Explanation:

The electric field inside a parallel plate capacitor is

$E = \frac{Q}{2\epsilon_0 A}$

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

$F = qE = \frac{qQ}{2\epsilon_0 A}$

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

$C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60$

Plugging this identity into the force equation above gives

$F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q$

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is $1.6 \times 10^{-19}$

Therefore, the kinetic energy is $2.4\times 10^{-19}$

8 0

3 years ago

A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During

Artemon [7]

Answer:

Induced EMF, $\epsilon=0.0143\ volts$

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, $B_i=200\ mT=0.2\ T$

Final magnetic field, $B_f=300\ mT=0.3\ T$

The expression for the induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}$

$\epsilon=0.0143\ volts$

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

5 0

3 years ago