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HACTEHA [7]
3 years ago
14

The ,______provides rigidity and protection to the plant cell​

Physics
1 answer:
balandron [24]3 years ago
7 0

Answer:

cell wall hope this will help

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A tray filled with ice is removed from the freezer. After a short period of time, the ice begins to melt.
Sav [38]
A: decreases in specific heat capacity
3 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

7 0
3 years ago
Why are house keys usually made of metal rather than glass
kherson [118]

Explanation:

doesn’t corrode easily and is soft enough for inexpensive tools to cut to the needed individual patterns.

4 0
3 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
3 years ago
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