All categories

3 years ago

A macro element that is a component of ATP

Physics

2 answers:

eimsori [14]3 years ago

7 0

Phosphorus is a macro element that is a component of ATP

ZanzabumX [31]3 years ago

5 0

Answer:

Phosphorus is a macro element that is a component of ATP.

Explanation:

You might be interested in

1. How much energy is needed to raise 1 g of water 1°C?

Serga [27]

Answer:

1.The calorie was originally defined as the amount of heat required at a pressure of 1 standard atmosphere to raise the temperature of 1 gram of water 1° Celsius. Since 1925 this calorie has been defined in terms of the joule, the definition since 1948 being that one calorie is equal to approximately 4.2 joules.

2.Boiling water at 100 degrees Celsius: 540 calories are needed to turn 1 gram (at 100 degrees Celsius) of water to steam.

5 0

4 years ago

Whu is it important to mave regular supervision of weights and measurements in the market

pentagon [3]

Answer:

Hope the answer helped you. if yes pls follow me

Explanation:

the fundamental answer is without regular supervision of definition of weights and measures,commerce exchange will be impossible and there would be no market whatsoever for anything.

5 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Gnesinka [82]

Answer:

a). $a_p=-2.39x10^{-12} rad/s^2$

b). $t=1016298.8 years$

c). $T_i=80.58x10^{-3}s$

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

$a_p=\frac{dw}{dt}$

$w=\frac{2\pi}{t}$

$a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}$

$dT=0.0808s$

$dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s$

Replacing

$a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2$

b).

If the pulsar will continue to decelerate at this rate, it will stop rotating at time:

$t=\frac{w}{a_p}$

$w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s$

$t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s$

$t=1016298.8 years$

c).

582 years ago to 2019

1437

$T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s$

5 0

4 years ago

One ball of mass 0.600kg travelling 9.00m/s to the right collides head on elastically with a second ball of mass 0.300kg travell

Alina [70]

Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

where v₁' and v₂' are the balls' respective velocities after the collision.

Kinetic energy is also conserved, so

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²

m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²

From the momentum equation, we have

(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'

which simplifies to

10.0 m/s = 2 v₁' + v₂'

so that

v₂' = 10.0 m/s - 2 v₁'

From the energy equation, we have

(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

which simplifies to

67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

226 m²/s² = 2 (v₁')² + (v₂')²

Substituting v₂' yields

226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²

which simplifies to

3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0

Solving for v₁' using the quadratic formula gives two solutions,

v₁' ≈ -2.33 m/s or v₁' = 9.00 m/s

but the second solution corresponds to the initial conditions, so we omit that one.

Then the second ball has velocity

v₂' = 10.0 m/s - 2 (-2.33 m/s)

v₂' ≈ 14.7 m/s

6 0

2 years ago

A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?

forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

$Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}$

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

$e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm$

Therefore, the extension of the wire is 0.362 mm.

8 0

3 years ago