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2 years ago

Do humans depend on photosynthesis? Explain how, why, or why not.

Physics

2 answers:

crimeas [40]2 years ago

5 0

Answer:

yes

Explanation:

just because we need it.

Elanso [62]2 years ago

4 0

Humans rely on photosynthesis to generate food as well as provide energy for heating, lighting, and electricity.

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The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

madam [21]

Answer:

(b) 1:1

Explanation:

1. Formulae:

E = hf

E = h.v/λ

λ = h/(mv)

E = (1/2)mv²

Where:

E = kinetic energy of the particle
λ = de-Broglie wavelength
m = mass of the particle
v = speed of the particle
h = Planck constant

2. Reasoning

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

$\dfrac{m_\alpha}{m_p}=4$

For the kinetic energies you find:

$\dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}$

$\dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4$

Thus:

$\dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}$

$m_\alpha v_\alpha=m_pv_p$

From de-Broglie equation, λ = h/(mv)

$\dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1$

5 0

3 years ago

shusha [124]

Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.

Explanation:

4 0

2 years ago

A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,

zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

4 0

3 years ago

A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of 2 Newton’s. What is the acce

Zina [86]

PHYSICS

Mass = 10 kg

Force = 2 N

Acceleration = ____?

Answers :

$f \: = m \times a$

2 = 10 × a

2 / 10 = a

0,2 m/s² = a ✅

7 0

3 years ago

According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind

Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0

3 years ago