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3 years ago

Which cell structure allows a plant to carry out photosynthesis?

Physics

2 answers:

NemiM [27]3 years ago

7 0

Answer: Chloroplasts allows plants to carry of photosynthesis.

kodGreya [7K]3 years ago

3 0

Answer:

Plant cells contain cell structures like cell wall, plastids, and large vacuoles. Cell wall provide plant cells rigidity and structural support and cell to cell interaction. Plastids help in storage of plant products. Chloroplasts aid in carrying out the process of photosynthesis to produce food for the plants.

Explanation:

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How would you write the number 6,500,000,000 in scientific notation?

photoshop1234 [79]

Scientific form = 6.5 x 109.

8 0

3 years ago

A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the

Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, $\Delta t=0.003\ s$

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

$F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N$

So, the average force on ball by the golf club is 340 N.

4 0

3 years ago

A simple pendulum is made by tying a 2.44 kg stone to a string 4.57 m long. The stone is projected perpendicularly to the string

alexdok [17]

Answer:

$v_{max}=8.2226m/s$

Explanation:

The problem is solved using the law of conservation of energy,

$mgL(1-cos\theta)+\frac{1}{2}mv^2_0=\frac{1}{2}mv^2_{max}$

$v_{max}=\sqrt{2gL(1-cos\theta)+v^2_0}$

$v_{max}=\sqrt{2(9.8)(4.57)(1-cos(69.4))+8^2}$

$v_{max}=8.2226m/s$

8 0

3 years ago

How much work is done in holding a 20 N sack of potatoes while waiting in line at the

Yuki888 [10]

Answer: A

Explanation:

honestly, it sounded the best

8 0

2 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers