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Zielflug [23.3K]
3 years ago
10

A 1.40-kg ball tied to a string fixed to the ceiling is pulled to one side by a force F→ . where L = 1.40 kg. What is the tensio

n in the string just before the ball is released and allowed to swing back and forth?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

T=13.72N

Explanation:

The tension before the ball is released have no angle is in rest at the same axis of the weight so:

∑F=0

Using Newton law in this case the ball is tied so tension before become to swing is

∑F=FN-T=0

T=F_{N}

T=m*g

T=1.40Kg*9.8\frac{m}{s^2}

T=13.72N

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10 points
MariettaO [177]

Answer:

true

Explanation:

yes because I said so

8 0
3 years ago
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
3 years ago
A 1022kg Caprice car stopped at an intersection is rear-ended by a 1620kg ranger truck moving with a speed of 14.5m/s. If the ca
Alika [10]

Answer:

Explanation:

mass of car, m = 1022 kg

mass of truck, M = 1620 kg

initial velocity of truck, U = 14.5 m/s

initial velocity of car, u = 0 m/s

Let the final velocity of car is v and the final velocity of truck is V.

Collision is elastic, so the coefficient of restitution, e = 1

Use conservation of momentum

initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck

m x u + M x U = m x v + M x V

0 + 1620 x 14.5 = 1022 v + 1620 V

23490 = 1022 v + 1620 V ..... (1)

Use the formula of coefficient of restitution

e = \frac{V_{1}-V_{2}}{u_{2}-u_{1}}

1 (14.5 - 0) = v - V

14.5 = v - V

V = v - 14.5 .... (2)

Put in equation (1)

23490 = 1022 v + 1620 (v - 14.5)

23490 = 1022 v + 1620 v - 23490

46980 = 2642 v

v = 17.8 m/s

Put in equation (2)

V = 17.8 - 14.5

V = 3.3 m/s

Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.

8 0
3 years ago
Determina la velocidad de un avión que recorre 459 km en 67 min
satela [25.4K]

Answer:

122397.7

Explanation:

4 0
3 years ago
What is the equivalent resistance of a circuit that contains four 75.0 resistors connected in parallel with a 100.0 V battery?
zhannawk [14.2K]
To get the total resistance in a parallel circuit, you need to remember that unlike in a series, you do not just merely add the resistances. You need to get the reciprocal first of each resistance and add them together. 

\frac{1}{R_{T}} =  \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}... +\frac{1}{R_{n}}

After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.

\frac{1}{R_{T}} = \frac{1}{75}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}

Add up the numerator and copy the denominator:

\frac{1}{R_{T}} = \frac{4}{75}

Then get the reciprocal to get the total resistance:

R _{T} = \frac{75}{4}  = 18.75

The answer to your question then is A. 18.8.
5 0
4 years ago
Read 2 more answers
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