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3 years ago

What protect and support the cell in a animal cell

Physics

2 answers:

DiKsa [7]3 years ago

6 0

In An animal cell the membrane protects it or also known as the cell wall.

sergey [27]3 years ago

4 0

The cell Membrane protects the cell in a animal cell

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2. What kinds of statistics are used in sports? In a paragraph explain your answer with examples

Lostsunrise [7]

Answer: Sports Illustrated noted that analytics in a sport such as a football is widely used to manage injury prevention. Basketball coaches make use of statistics such as field-goal attempts, effective field goal percentage, free throw attempts, and free throw percentages.

Explanation: BRAINLEST?

4 0

2 years ago

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
where the force is F=qE, d=0.556 and $\theta=55.2^{\circ}$ . Using the value of q and E given by the problem, we find
$W=qEdcos\theta = 6.39\cdot10^{-5}J$

3 0

3 years ago

The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the

natta225 [31]

Answer:

50 N

Explanation:

Efficiency of a machine can't be more than 1, so I assume you mean 40%. (Remember, efficiency and mechanical advantage are not the same).

Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

We know that e = 0.40, and Fout = 120 N. Since there are 6 pulleys, we also know that Din/Dout = 6.

F = (120 N / 0.4) / 6

F = 50 N

5 0

2 years ago

One event occurs at the origin at t equal to zero, and a second events occurs at the point x=5m along the x-axis at time with ct

Anastasy [175]

Answer:

The separation will be spacelike.

The first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

Explanation:

We can define the separation between two events (using the + - - - signature) as :

$(\Delta s )^2 = (ct_2 - c t_1 )^2 - (x_2 - x_1)^2$

where the separation will be lightlike if is equal to zero, timelike if is positive and spacelike if is negative.

For our problem

$c t_1 = 0$

$x_1 = 0$

$ct_2 = 4 \ m$

$x_2 = 5 \ m$

$(\Delta s )^2 = (4 \ m - 0 )^2 - ( 5 \ m - 0)^2$

$(\Delta s )^2 = (4 \ m )^2 - ( 5 \ m 0)^2$

$(\Delta s )^2 = 16 \ m^ 2 - 25 \ m^2$

$(\Delta s )^2 = - 9\ m^2$

So the separation will be spacelike, and the first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

8 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago