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3 years ago

HELP ASAP!!!!!!! PLEASE WILL GIVE BRAINLIEST

1 answer:

7 0

Answer: The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. Not all molecules are free to diffuse. If plasma membrane ruptures or breaks down then molecules of some substances will freely move in and out of the cells.

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Penguin #1 travels at speed v=0.9 c relative to penguin #2. At the instant that both are at the origin, they synchronize their c

Alekssandra [29.7K]

Answer:

L = 0.44 [m]

Explanation:

Here we can use the Lorentz transformation related to length to solve it:

$L=L_{0}\sqrt{1-\beta^{2}}$

Where:

L₀ is the length of the moving reference frame (penguin #1)

L is the length of the fixed reference frame (penguin #2)

β is the ratio between v and c

We know that v = 0.9c so we can find β.

$\beta = \frac{0.9c}{c}=0.9$

$L=1 [m]\sqrt {1-0.9^{2}} = 0.44 [m]$

Therefore, the length of the meter stick of #1 observed by #2 is 0.44 m.

I hope it helps you!

6 0

4 years ago

More free stuff have fun

marusya05 [52]

Happy hanukkah or christmas people

6 0

3 years ago

Read 2 more answers

In a carnival game, the player throws a ball at a haystack. For a typical throw, the ball leaves the hay with a speed exactly on

8_murik_8 [283]

Answer:

Ve(m) = sqrt (19.2/m)

Ve(0.35) = 7.407 m/s

Explanation:

Given:

- The ball has a mass = m

- The entry speed of the ball is Vi = Ve

- The final speed of the ball Vf = 0.5*Ve

- The constant frictional force on ball due to hay is F = 6 N

- The thickness of hay-stack is s = 1.2 m

- Assume the throw is in horizontal direction and neglect gravity forces

Find:

Derive an expression for the typical entry speed as a function of the inertia of the ball

What is the typical entry speed if the ball has an inertia of a 0.35 kg?

Solution:

- To determine the entry speed as a function of inertia we will use third equation of motion as follows:

Vf^2 = Vi^2 + 2*a*s

Where, a is acceleration of the ball through hay stack. We will use Newton's Law of motion to determine this:

F_net = m*a

The only force acting on the ball in its journey through hay-stack is the frictional force F:

- F = m*a

a = -F/m

- Input all the quantities in the third equation of motion:

(0.5Ve)^2 = Ve^2 - 2*F*s / m

0.75Ve^2 = 2*F*s / m

Ve = sqrt (8*F*s/3*m)

Plug in values:

Ve(m) = sqrt (8*6*1.2/3*m)

Ve(m) = sqrt (19.2/m)

- The entry speed for the inertia of the ball m = 0.35 kg is:

Ve(0.35) = sqrt(19.2/0.35)

Ve(0.35) = 7.407 m/s

8 0

4 years ago

A string of length L= 1.2 m and mass m = 20g is under 400 N of tension, its two ends are fixed. a. How many nodes will you see i

ozzi

Answer:

(iii) 6

Explanation:

Part a)

Since it is given that both ends are fixed and it is vibrating in 5th harmonic

So here it will have 5 number of loops in it

so we can draw it in following way

each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes

So there are 5 loops which means it will have 5 antinodes

and hence there will be 6 nodes in it

so correct answer will be

(iii) 6

8 0

3 years ago

you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t

telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

Your question is not complete, it seems to be missing the following information;

"The velocity of the ant is 2 m/s"

The given parameters;

velocity of the ant, v = 2 m/s

change in position of the ant, Δx = 0.81 m

initial time, t₁ = 4 s

time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

$v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s$

The time elapsed since you stopped the stopwatch is calculated as;

$t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s$

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

5 0

3 years ago