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Allushta [10]
3 years ago
9

Mercury is characterized by

Physics
1 answer:
cupoosta [38]3 years ago
8 0
 <span>It's close to the sun without much atmosphere, so it's characterized by </span><span>very extreme temperatures.

Happy studying ^_^</span>
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Which of these help create radio waves?
Margaret [11]
The correct option is A.
All waves are divided into two basic classes, which are transverse and longitudinal waves. A transverse wave is a type of wave in which the medium of propagation vibrates at angle 90 degree in relation to the direction of propagation. Radio wave is an example of a transverse wave. 
5 0
3 years ago
Read 2 more answers
My Notes A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field t
Nastasia [14]

Answer:

0.00158 A

Explanation:

A = Area of loop = 7.8 cm²

B_f = Final magnetic field = 3.2 T

B_i = Initial magnetic field = 0.5 T

R = Resitance = 1.3\ \Omega

The emf is given by

E=\frac{(B_f-B_i)A}{t}\\\Rightarrow E=\frac{(3.2-0.5)\times 7.8\times 10^{-4}}{1.02}\\\Rightarrow E=0.00206\ V

The current is given by

I=\frac{E}{R}\\\Rightarrow I=\frac{0.00206}{1.3}\\\Rightarrow I=0.00158\ A

The resulting induced current in the loop is 0.00158 A

6 0
4 years ago
PLEASE PLEASE PLEASE HELP ME!!!!​
Firlakuza [10]

31) The head of each arrow represents the direction of each force

32) The length of each arrow represents the magnitude of each force

33) The net  force on the 1-kg object is to the left

34) To balance the forces on the object, we must add a force acting to the right

35) The acceleration of the 1-kg object is twice the acceleration of the 2-kg object

36) The magnitude of the acceleration increases, and the direction changes

Explanation:

31-32)

A free-body diagram is a diagram representing the forces acting on a body.

In a free-body diagram, each force is represented with an arrow, where:

  • The length of the arrow is proportional to the magnitude of the force
  • The direction of the arrow represents the direction of the force

Therefore:

- The head of each arrow represents the direction of each force

- The length of each arrow represents the magnitude of each force

33)

In the free-body diagram of the 1-kg object, we observe that the arrow on the right is longer than the arrow on the left. This means that the force acting from the right (and therefore, acting TO the left) is stronger than the force acting from the left (which is therefore acting TO the right).

Therefore, the net force acting on the 1-kg object is to the left, since the force acting to the left is stronger.

34)

In order to balance the forces acting on the 1-kg, we should apply an additional force such that the net force acting on the object is zero.

In part 33), we said that the net force acting on the 1-kg object is to the left. Therefore, in order to produce a net force of zero, we must add another force acting to the right, in order to balance the current net force pointing to the left.

35)

First of all, by looking at the two free-body diagrams, we notice that the magnitudes of the forces acting on the 1-kg and 2-kg objects are the same. This means that the net force acting on the two objects is the same.

Now we can apply Newton's second law of motion to the two cases; the law states that the net force acting on an object is equal to the product between its mass and its acceleration:

\sum F = ma

where \sum F is the net force, m is the mass, a is the acceleration

For the object of 1-kg, we have:

\sum F = m_1 a_1

where m_1 = 1 kg

For the object of 2-kg, we have:

\sum F = m_2 a_2

where m_2 = 2 kg

We said that the net force on the two objects is the same, so we can combine the two equations:

m_1 a_1 = m_2 a_2\\(1 kg)a_1 = (2 kg)a_2\\\rightarrow a_1 = 2a_2

Therefore, the acceleration of the 1-kg object is twice the acceleration of the 2-kg object.

36)

If an upward force is applied to the 1-kg object, the acceleration will increase in magnitude and it will change in direction.

In fact, let's call F_x the total net force acting in the horizontal direction on the 1-kg object. Let's call F_y the upward additional force applied. This means that now the object will have two components of the acceleration:

a_x, horizontally (to the left)

a_y, vertically (upward)

So, the magnitude of the net acceleration will be

a=\sqrt{a_x^2+a_y^2}

which is larger than the original acceleration (a_x). Also, since there is a component of the acceleration in the vertical direction, the final direction of the acceleration a will be between left and upward.

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

4 0
4 years ago
A 200-kg boulder is 1000-m above the ground.
dem82 [27]

Answer: 1,960,000 J

Explanation:

7 0
3 years ago
Jumps at 8 m/s at a 30 degree angle above the horizontal. how long was she in the air?
pentagon [3]
Vo = 8 m/s. g = 10 m/s^2
alpha = 30 degree
ask: x?

x = vo^2 sin ^2 alpha / g
x = 8 ^2 x 2 sin 30 cos 30 / 10
x =( 64 x2x 1/2x1/2√3 )/ 10
x= 3.2 √3 metres

√3 = 1.73
4 0
3 years ago
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