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1 year ago

During photosynthesis, plants transform the energy in sunlight into chemical

Physics

1 answer:

shutvik [7]1 year ago

3 0

B and D

Explanation:

The scientists need to replicate her experiment and to do that you need actual sunlight,it can't be carried out in a laboratory

Hope this helped

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Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point

Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

$B = \frac{\mu_oI}{2\pi R}$

At the center, the magnetic field strength is twice

$B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}$

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

$B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A$

Therefore, current needed is 39.8 A

6 0

2 years ago

Two forces F1 = -7.90i + 5.40j and F2 = 9.00i + 7.60j are acting on an object with a mass of m = 6.30 kg. The forces are measure

IceJOKER [234]

Answer:

Propotionality is important

Explanation:

7 0

1 year ago

A bowling ball is rolling in a straight line toward the pins at a speed of 2.2 m/s. Its momentum is 4.75 kg*m/s. What is the

Thepotemich [5.8K]

Answer: m= 2.16 kg

Explanation: Momentum is expressed in the following formula:

p = mv

Derive to find m:

m = p / v

= 4.75 kg.m/s / 2.2 m/s

= 2.16 kg

Cancel out m/s and the remaining unit is in kg.

3 0

3 years ago

Read 2 more answers

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0

1 year ago

What occurs when a swimmer pushes through water to swim

Triss [41]

When we swim we apply force and push the water backward with the help of our hands. In response, The water pushes us forward with an equal force. Thus, in order to move forward and swim, the swimmer lushes the water backward. Newton's 3rd law of motion

6 0

2 years ago