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goldfiish [28.3K]
3 years ago
14

What part of a plant cell traps sunlight?_____________________

Physics
2 answers:
elixir [45]3 years ago
4 0
It is the <span>Chloroplast in the cells.</span>
sineoko [7]3 years ago
4 0
The green pigment that traps<span> energy from </span>sunlight<span> is called chlorophyll. This pigment is found in special </span>cell<span> organelles called chloroplasts.
</span><span>
Its Chlorophyll</span>
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What current flows through a 2.56-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 ✕ 103 V is applied to it? (Suc
vfiekz [6]

Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, \rho=2300\ \Omega-m

Length of rod, l = 20 cm = 0.2 m

Voltage, V=1\times 10^3\ V

The resistivity of the rod is given by :

R=\rho\dfrac{L}{A}

R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

I=\dfrac{V}{R}

I=\dfrac{10^3\ V}{893692.30\ \Omega}

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

6 0
2 years ago
Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
Alexxx [7]

Answer:

1).atoms (3). mixture. (5). Element

2). particles (4). molecules (6). suspension

Explanation:

(7). Homogeneous (8). Heterogeneous

(9). compound (10). solutions

8 0
3 years ago
A heavy truck and a small truck roll down a hill. Neglecting friction, at the bottom of the hill the heavy truck has greater
postnew [5]

Answer:

kenetic energy

Explanation:

or potential energy

3 0
3 years ago
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
2 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
2 years ago
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