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dem82 [27]
4 years ago
12

Match the followings.

Physics
1 answer:
Whitepunk [10]4 years ago
3 0

Answer:

a. The current is induced when there is ____3. EITHER DECREASING OR INCREASING___ magnetic flux through a closed loop of wire.

b. If the magnetic flux was constant then there _______WILL BE NO___ induced current regardless of the magnetic flux value.

c. If the magnetic flux was not constant then there __WILL BE______ induced current regardless of the magnetic flux value.

Explanation:

THIS IS BECAUSE IF FARADAY'S LAW OF ELECTROMAGNETIC INDUCTION WHICH STATES THAT WHENEVER THERE IS A CHANGE IN MAGNETIC LINES OF FORCE LINKED WITH A CLOSED CIRCUIT AN EMF IS INDUCED

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Can someone help me because I don’t really understand thxxx
almond37 [142]

F=ma

100N=50kg*a

a=100N/50kg=2m/s^2

8 0
3 years ago
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Obj
Anna35 [415]

Answer:

<em>The object with the twice the area of the other object, will have the larger drag coefficient.</em>

<em></em>

Explanation:

The equation for drag force is given as

F_{D} = \frac{1}{2}pu^{2}  C_{D} A

where F_{D} IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

C_{D} = coefficient of drag

A = area of the object

Note that C_{D} is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

F_{D} ∝ P_{D} A

where P_{D} is the pressure exerted by the fluid on the area A

Also note that P_{D} = \frac{1}{2}pu^{2}

which also clarifies that the drag force is approximately proportional to the abject's area.

<em>In this case, the object with the twice the area of the other object, will have the larger drag coefficient.</em>

5 0
3 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
4 years ago
Which two statements describe how ultrasound technology produces an image of a baby before it is born?
Elan Coil [88]

Answer:a and c

Explanation:

8 0
3 years ago
Read 2 more answers
What is depth perception?
rjkz [21]
Depth perception is the way in which some animals, such as humans, can tell if a object is close or far away from them.
6 0
3 years ago
Read 2 more answers
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