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iren2701 [21]
2 years ago
9

This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Ear

th without flying off or hitting the ground.​

Physics
1 answer:
LenaWriter [7]2 years ago
6 0

Answer:

One force will be gravity & inertia.

Explanation:

Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth

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If the car went 30 km west in 25 min. and then 40 km east in 35 min., what would be its total displacement?
Mandarinka [93]
The total displacement is 10 km.
8 0
3 years ago
A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri
lys-0071 [83]
The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
                        P = W = mass x acceleration due to gravity
    
                    P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N

Solving for the static friction force (F), 
                                              F = P x c 
 
                                      F = (2.94 N) x 0.6 = 1.794 N

Therefore, the maximum force of static friction is 1.794 N. 



5 0
2 years ago
Read 2 more answers
If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step
azamat

Answer:

Explanation:

v_f^2 = v_i^2-2a(x_f-x_i)

Subtract both sides by v_i^2:

- v_i^2+v_f^2 = -2a(x_f-x_i)

Divide both sides by -2*a:

\frac{v_i^2 - v_f^2}{2a} =x_f-x_i

Add both sides by x_i:

x_i+\frac{v_i^2 - v_f^2}{2a} =x_f

Subtract both sides by \frac{v_i^2 - v_f^2}{2a}:

x_i=x_f-\frac{v_i^2 - v_f^2}{2a}

8 0
3 years ago
Explain how polarization of a cell increases the cell's internal resistance.<br>(2<br>2.​
Mandarinka [93]

Answer:

Explanation: The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.

   A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.

   One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.

   A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.

LOCAL ACTION

   When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.

   Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.

6 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
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