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blsea [12.9K]
3 years ago
9

Tim kept two identical green plants in his garden and labeled them A and B. He watered plant A every day and plant B once a mont

h. After a month, the leaves of plant A remained green but the leaves of plant B turned yellow. Which statement best explains this difference in color? Plant A got sufficient water to produce chlorophyll. Plant A got sufficient water to produce carbon dioxide. Plant B did not get sufficient water to convert carbon dioxide into hydrogen. Plant B did not get sufficient water to convert carbon dioxide into glucose.
Physics
2 answers:
valentinak56 [21]3 years ago
7 0

Answer: Correct answer on Quiz is

Plant B did not get sufficient water to convert carbon dioxide into glucose.

Explanation:

tino4ka555 [31]3 years ago
3 0
The answer is Plant A got sufficient water to produce chlorophyll.  Tim kept two identical green plants in his garden and labeled them A and B. He watered plant A every day and plant B once a month. After a month, the leaves of plant A remained green but the leaves of plant B turned yellow.  The best statement that explains the difference in color is that Plant A got sufficient water to produce chlorophyll.
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Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) 0.080, (b) 0.26, (c) 0.49, (d) 0.90, (e)
mote1985 [20]

Answer:

Explanation:

This is the case of forced oscillation . The pendulum having the same or matching time period  or angular frequency with that of angular frequency  of external periodic force , will be in resonance having largest amplitude.

Angular frequency of pendulum having length .9 m

= \sqrt{\frac{g}{l } }

l = .9

angular frequency

= \sqrt{\frac{10}{0.9 } }

= 3.33 rad / s

If we calculate angular frequencies of pendulum of all lengths given , we will find that other lengths do not give angular frequency falling between 2 and 4 radian . So only pendulum having length of .9 m will have vibration of maximum amplitude.

8 0
3 years ago
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
NEED HELP ASAP WILL MARK BRAINLIEST.
-Dominant- [34]
G is the answer for apex vs / Chehhhh
3 0
3 years ago
A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
3 years ago
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate
LenKa [72]

Answer:

12.64968 Hz

Explanation:

v = Velocity of sound in seawater = 1522 m/s

u = Velocity of dolphin = 7.2 m/s

f' = Actual frequency = 2674 Hz

From Doppler effect we get the relation

f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz

The frequency that will be received is 2661.35032 Hz

The difference in the frequency will be

2674-2661.35032=12.64968\ Hz

6 0
4 years ago
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