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3 years ago

10.

Physics

1 answer:

nlexa [21]3 years ago

3 0

Answer:

Cell Membrane

Explanation:

The cell membrane controls what goes in and out of a cell, and keeps it shape, much like a city limit.

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Please help will be marked most brainlist !!!

Neko [114]

Answer:

Explanation:

all of the rest are true

6 0

2 years ago

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu

ella [17]

Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

Explanation:

The given spring constant of the of the spring, k = 88.0 N/m

The length by which the hose is stretched, x = 4.20 m

For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose

The elastic potential energy, P.E., of a compressed spring is given as follows;

P.E. = 1/2·k·x²

∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²

1/2 × 88.0 N/m × (4.20 m)² = 776.16 J

The work done on the hose = The potential energy given to hose, P.E. = 776.16 J

5 0

3 years ago

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

6 0

3 years ago

What cause earthquakes?

KengaRu [80]

Question; what causes earthquakes?
Answer; what causes earthquakes are mainly when rocks are underground and then suddenly break along a fault.

6 0

3 years ago

Read 2 more answers

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$

$\theta = tan^{-1} (\frac{d}{\mu L})$

Replacing,

$\theta = tan^{-1} (\frac{0.9}{0.42*2})$

$\theta = 46.975\°$

Therefore the minimum angle that the person can reach is 46.9°

8 0

3 years ago