Necesito ayudaaaaaa por favor

Which type of muscle cell can have multiple nuclei

s344n2d4d5 [400]

Answer:

Skeletal muscle cells

Explanation:

Skeletal muscle cells are long, cylindrical, and striated. They are multi-nucleated meaning that they have more than one nucleus. This is because they are formed from the fusion of embryonic myoblasts.

pls mark me the brainliest

7 0

2 years ago

Read 2 more answers

The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va

Hitman42 [59]

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

5 0

2 years ago

A 5.0 kg object moving at 10 m/s on a frictionless surfaces collides with but does not stick to a 2.0 kg object that is initiall

lilavasa [31]

Answer:

I= 20 i {N.s}

Explanation:

In order to obtain the impulse on the 2 kg ball, you have to apply the equation of Impulse:

I=FΔt

Where I is the impulse vector, F is the net force and Δt is the interval of time when the force is applied.

In this case:

Δt=0.01 s

F= 2000 i N

where i is the unit vector in the x direction.

Replacing the values in the formula:

I=(2000)(0.01)i

Therefore:

I= 20 i {N.s}

3 0

3 years ago

A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string,

Lorico [155]

Answer:

v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

y = y₀ + v_{oy} t - ½ g t²

0 =y₀ + 0 -1/2 gt²

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 40/32)

t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

sin θ = y / l

θ = sin⁻¹1 y / l

θ = sin⁻¹ 40/50

θ = 53.1º

therefore the horizontal distance is

x = l cos 53.1

x = 50cos 53.1

x = 30 m

let's use the equation

x = v₀ₓ t

v₀ₓ = x / t

v₀ₓ = 30 / 2.5

v₀ₓ = 12 m / s

we look for the vertical component of the velocity

v_y = v_{oy} - g t

v_y = 0 - g t

v_y = - 9.8 2.5

v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

v = $\sqrt{v_x^2 + v_y^2}$

v = $\sqrt{12^2 + 24.5^2}$

v = 27.28 m /s

we use trigonometry for the angle

tan θ = v_y / vₓ

θ = tan⁻¹ v_y / vₓ

θ = tan⁻¹ 24.5 / 12

θ = 63.9º

4 0

2 years ago

A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300

Illusion [34]

<h2>Answer:</h2>

$\boxed{P=96.09MW}$

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

$R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega$

So we can calculate the power loss as:

$P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}$

Finally, the power loss due to resistance in the line is 96.09MW

5 0

3 years ago