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Masja [62]
3 years ago
14

There is a species of bamboo that can grow 36 inches per day. If a plant grew at this rate and was measured at 40 inches initial

ly, how long (in days) would it take for the plant to grow to a height of 200 inches tall?
Physics
1 answer:
LekaFEV [45]3 years ago
8 0

Answer:

It will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

Explanation:

From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.

To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.

This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.

Now,

If the plant grows 36 inches in 1 day

then it will grow 160 inches in x days

x = (160 inches × 1 day) / 36 inches

x = 160 / 36

x = 4\frac{4}{9} days or 4.44 days

Hence, it will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

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NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the ener
Stella [2.4K]

Answer:

Reflective

Explanation:

The radiation pressure of the wave that totally absorbed is given by;

P_{abs}= \frac{I}{C}

and While the radiation pressure of the wave totally reflected is given by;

P_{ref}= \frac{2I}{C}

Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.

6 0
3 years ago
Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If
cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

7 0
3 years ago
According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.
nignag [31]

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

\epsilon = -L\frac{dI}{dt}

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V

Therefore, the self-induced emf in this inductor is 4.68 mV.  

I hope it helps you!

6 0
3 years ago
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