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natita [175]
3 years ago
9

Which of the following statements correctly describes the role that friction plays in the motion of a train? *

Physics
2 answers:
Anna007 [38]3 years ago
8 0
C. FRICTION IS NECESSARY TO KEEP THE TRAIN ON THE TRACK AS IT IS MOVING
Lesechka [4]3 years ago
7 0

Answer:

Should be D!

Explanation:

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Fatigue strength is generally significantly improved by using high steel a. alloy b. yield c. hardened d. ultimate strength e. a
Gala2k [10]

Answer:

e. all of these

Explanation:

The fatigue strength is improved by then high alloy steels , high yield steels , high hardened steel , high ultimate steel .

Due to the formation of the improved materials in alloy steels will increase the fatigue strength . Similarly for a high yield steels and hardened steels these cycles to failure will improve .

7 0
3 years ago
Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 1
Novosadov [1.4K]
<span>they are travelling at right angles to each other.
 At any given instant they form a right triangle with their starting point
 </span>South bound <span>= x  [mi/h]
</span> East bound <span> = x+1 [mi/h]
 after five hours they will be
 d=5x
 and
 d=5(x+1)
 miles away from the starting point 
 (5x)^2+(5(x+1))^2=625
 25x^2+(5x+5)^2=625
 25x^2+25x^2+50x+25=625
 50</span>x^2+50x-600=0
<span> x^2+ x - 12=0
 (x+4)(x-3)=0
 take the postive value
 x= 3 mph the speed of south bound
 4mph east bound </span>
6 0
3 years ago
An 8 Ω resistor is connected to a 6 V battery and another resistor in series. If the current in the circuit is 0.5 A, what is th
Semmy [17]
8+R=V/I
8+R=6/0.5=12 ohm
then R=4 ohm
4 0
3 years ago
The ears are ____ and ____ to the shoulders and ____ to the nose.
Levart [38]
The ears are superior and posterior to the shoulders and lateral to the nose.
3 0
3 years ago
Read 2 more answers
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
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