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4 years ago

What is the main function of chloroplasts?

Physics

1 answer:

dangina [55]4 years ago

6 0

Answer:

To convert sunlight into usable sugars

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Explanation:

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Specific heat is the amount of heat required to raise the temperature of a material by 1°C per what unit?

nydimaria [60]

The heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

or C. Mass if you're on plato

3 0

3 years ago

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A spring stretches by 0.014 m when a 3.7-kg object is suspended from its end. How much mass should be attached to this spring so

lidiya [134]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

3 0

3 years ago

a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v

shutvik [7]

Isothermal Work = PVln(v₂/v₁)

PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = $e^{0.3098}$

v₂ = 19* $e^{0.3098}$

v₂ = 25.8999

v₂ ≈ 26 L Option b.

6 0

3 years ago

What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the

alisha [4.7K]

Answer:

1.1648×10⁻¹¹ N

Explanation:

Using

F = qvBsinФ..................... Equation 1

Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.

Note: The angle between v and B = 90°

Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°

Substitute into equation 1

F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°

F = 11.648×10⁻¹²

F = 1.1648×10⁻¹¹ N.

6 0

3 years ago

Read 2 more answers