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ddd [48]
3 years ago
14

A scientist measures the growth of a bamboo plant over time. The table above shows the results. What is the best interference fo

r the height of the plant after 45 hours?
A. 2.40 cm
B. 2.25 cm
C. 2.55 cm
D. 2.10 cm​

Physics
1 answer:
lisabon 2012 [21]3 years ago
7 0

(B) 2.25cm

<u>Explanation:</u>

Given:

At 40 hours, the height of the bamboo plant is 2.1cm

At 50 hours, the height of the bamboo plant is 2.4cm

Height of the bamboo plant after 45 hours = ?

The difference in length from 40 to 50 hours = 2.4 - 2.1cm

                                                                      = 0.3 cm

Mean of 40 and 50 is 45.

Thus,

At 45 hours, the height will increase by 0.3/2

                                                         = 0.15 cm

Height at 45 hour = 2.1 + 0.15cm

                           = 2.25cm

Therefore, the height of the plant after 45 hours is 2.25cm

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Furkat [3]
            <span> Using conservation of energy

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4 0
2 years ago
Read 2 more answers
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

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3 0
3 years ago
A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the
Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

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Mgh = \frac{1}{2}IW^2 +\frac{1}{2}MV^2

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

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Also, the moment of inertia of the disk is equal to:

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I = \frac{1}{2}(5kg)(2m)^2

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so, we can write the initial equation as:

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V = 2 m/s

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