Pls help i need an answer for this ill mark u brianlist or anything u want

Describe how molecules enter and leave a cell without the use of the cellâ€™s energy

HACTEHA [7]

Answer:

Through passive transport

Explanation:

<em>Passive transport is the movement of molecules and ions across the cell membrane whith no use of energy,</em> it usuale involves a difussion down its concentration gradient (a region where concentration changes) across a membrane, from higher to lower concentrations. Passive transport can happen by simple diffusion, facilitated diffusion, filtration, or osmosis.

I hope you find this information useful and interesting! Good luck!

5 0

4 years ago

If a sound wave is produced with a wavelength of 1.04m what is the waves frequency

dmitriy555 [2]

You should just ask the wave

5 0

3 years ago

A spring (k=15.19kN/m)is is compresses 25cm and held in place on a 36.87° incline. A block (M=10kg) is placed on the spring. Whe

Savatey [412]

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

$E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1$

Where $U_e_0$ is the elastic potential energy stored in the spring, $K_1$ is the kinetic energy of the block at point 1, $U_g_1$ is the gravitational potential energy of the block at point 1, and $W_f_1$ is the work done by friction at point 1.

Now, rearranging the equation we obtain:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1$

Where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $v_1$ is the speed at point 1, $g$ is the acceleration due to gravity, $h_1$ is the vertical height of the block at point 1, $\mu$ is the coefficient of kinetic friction, $N$ is the magnitude of the normal force and $s_1$ is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

$N=mg\cos\theta$

Where $\theta$ is the angle of the ramp.

We can find the height $h_1$ using trigonometry:

$h_1=s_1\sin\theta$

Then, our equation becomes:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}$

Plugging in the known values, we get:

$v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s$

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

$v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s$