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1 year ago

The long structure of small intestine is accommodated in small space within our body. Comment.

1 answer:

6 0

The long structure of small intestine is accommodated in small space within our body because of extensive coiling. the small intestine is highly coiled structure and thus can easily be fixed in a small space.

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A 6.0 kg metal ball moving at 4.0 m/s hits a 6.0 kg ball of putty at rest and sticks to it. The two go on at 2.0 m/s.

lutik1710 [3]

The metal ball lost energy while the putty ball gained energy.

<h3>What is momentum?</h3>

Momentum is the product of mass and velocity of the body. We must note that momentum before collision is equal to momentum after collision.

1) Kinetic energy before collision = 1/2mv^2 = 0.5 * 6 * 4 = 12 J

2) kinetic energy after collision = 0.5 * 6 * 2= 6 J

3) Kinetic energy of putty ball = 0.5 * 6 * 2= 6 J

4) Energy lost by the metal ball = 12 J - 6 J = 6 J

5) Energy gained by the putty ball = 6 J - 0J = 6 J

6) The rest of the energy was converted to heat after the collision.

Learn more about kinetic energy: brainly.com/question/999862

3 0

2 years ago

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

3 years ago

Describe what a hydrogen bond is. 6th grade answer

r-ruslan [8.4K]

Answer:

<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>

Explanation:

8 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago

Please Someone Help Me...

goldenfox [79]

Answer:

Hi myself Shrushtee.

Explanation:

The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current. Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire.

6 0

2 years ago