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Artyom0805 [142]
4 years ago
10

U

Physics
1 answer:
katen-ka-za [31]4 years ago
8 0

Answer: B = 0.1 Tesla

Explanation: Given that the

Emf = 6.0 V

Length L = 30 m

Speed V = 2 m/s

B = ?

The induced EMF = BVL

6 = 30 × 2 × B

B = 6/60

B = 0.1 Tesla

The magnetic field B = 0.1 Tesla.

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A car traveling west in a straight line on a highway decreases its speed from 30.0 meters per second to 23.0 meters per second i
pychu [463]
Acceleration, a =  (v - u) / t

Initial Velocity, u = 30 m/s
Final Velocity, v  =  23 m/s
time                 t  =  2.00 seconds

a = (23 - 30) / 2
a = -7 / 2 = -3.5 m/s2

So the acceleration  is negative, which means it is a deceleration of 3.5 m/s2.
8 0
3 years ago
A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a ve
SIZIF [17.4K]

90 degrees - 30 = 60 degrees

Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)

Velocity = 2.47 m/s

The answer is D) 2.47 m/s at 61.9 degrees

8 0
3 years ago
7) Germanium, element 32 on the Periodic Table, is shown here. If a proton is added to the nucleus of germanium, what outcome(s)
cestrela7 [59]

Addition of a proton to germanium  will convert it to arsenic (element 33) having different properties.

The atomic number of an atom is the number of protons in the nucleus of the atom. The atomic number serves as the identity of an atom. If the atomic number is changed by adding or removing protons, the identity of that atom changes.

Hence, when a proton is added to germanium, the  atom would become arsenic (element 33) and have different properties.

Learn more: brainly.com/question/14281129

5 0
3 years ago
A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from
Rzqust [24]

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

Learn more about brainly.com/question/874205 here:

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7 0
2 years ago
Read 2 more answers
Identify and explain which ball most likely had the greatest speed.
deff fn [24]
Y, bc the height of the bounce back is higher than x
4 0
3 years ago
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