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3 years ago

5

For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 79 kg adult receive

?

1 answer:

Gennadij [26K]3 years ago

5 0

12 mg = 0.012 g
12 mg = 1.2e-5 kg

75 kg = 75,000 g

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Ronch [10]

Answer:

Current in a parallel circuit = 0.61 amps (Approx)

Explanation:

Given:

Voltage V = 6 volt

Two resistors = 17.2 , 22.4 in parallel circuit

Find:

Current in a parallel circuit

Computation:

1/R = 1/r1 + 1 / r2

1/R = 1/17.2 + 1 / 22.4

R = 9.73 ohms (Approx)

Current in a parallel circuit = V / R

Current in a parallel circuit = 6 / 9.73

Current in a parallel circuit = 0.61 amps (Approx)

4 0

3 years ago

A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potentia

aivan3 [116]

Answer:

$\Delta V=V_{2}-V_{1}=45.4V$

Explanation:

The energy, E, from a capacitor, with capacitance, C, and voltage V is:

$E=\frac{1}{2} CV^{2}$

$V=\sqrt{2E/C}$

If we increase the Voltage, the Energy increase also:

$V_{1}=\sqrt{2E_{1}/C}$

$V_{2}=\sqrt{2E_{2}/C}$

The voltage difference:

$V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}$

$V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V$

5 0

3 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

brainly.com/question/7973509

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#LearnwithBrainly

7 0

3 years ago

Internal diameter of a test tube<br>

sladkih [1.3K]

Answer:

16 mm with 14 mm inside diameter.

4 0

3 years ago

A large ball of Play-Doh with a mass of 0.04kg is launched from a catapult with an initial

ryzh [129]

Answer:

≈ 6.68 m/s

Explanation:

A suitable formula is ...

vf^2 -vi^2 = 2ad

where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.

We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.

Then ...

vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s

vf ≈ 6.68 m/s

The ball hits the ground with a speed of about 6.68 meters per second.

__

We assume the launch direction is either up or down.

3 0

3 years ago