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Ivahew
1 month ago
8

According to the first rule, if a force pulls on one end of a rope, the tension in the rope equals the magnitude of the pulling

force. What is the magnitude T1 of the tension in rope 1? Express your answer in terms of some or all of the variables F, mB and g.

Physics
1 answer:
Sav [3.1K]1 month ago
7 0

Answer:

The provided documents adequately explain the situation. We'll identify the force exerted on the rope using Newton's third law, which underscores that this problem relates to the concept of equilibrium.

The outcome reveals that "The tension T1 in rope 1 matches the force F.

Explanation:

According to Newton's third law, for every action, there is an equal and opposite reaction.

Refer to the documents for a thorough solution and clarification of the issue.

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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
Ostrovityanka [3204]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time t_{sound}. Thus, we can derive:

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

t_{arrow} + t _{sound} = 1 s.

Using this relationship in the distance formula for sound allows us to write:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Substituting the value of d from the first equation yields:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, after some calculations, we can proceed further:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Finally, the value is inserted into the initial equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
28 days ago
An electron is projected with an initial speed of 3.9 × 105 m/s directly toward a proton that is fixed in place. If the electron
Keith_Richards [3271]

Explanation:

Attached is a document that contains the solution.

8 0
14 days ago
These little bunnies (A, B, and C) were born in the same litter to the same parents, but they have different traits for the feat
Ostrovityanka [3204]

Response/Clarification:

Each of us receives 2 versions of a gene from our parents, with one inherited from the mother and one from the father.

Both our mother and father possess 2 versions of every gene. Therefore, the specific version we inherit is determined randomly, much like the outcome of a coin flip. This applies to both parents.

For instance, if the mother has one variant causing thick ears (A) and another for thin ears (a), she is Aa.

Similarly, the father also has these variants and is Aa as well.

The father can transfer either A or a, while the mother can also transfer either A or a.

As a result, their offspring can be AA, Aa, or aa. An AA genotype results in thick ears, while aa results in thin ears. The Aa genotype produces ears of intermediate thickness, akin to bunny B. This demonstrates the concept of incomplete dominance

8 0
14 days ago
The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2
serg [3582]

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!

0 0
2 months ago
Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
Keith_Richards [3271]

The answer is:

V=14m/s

Details are as follows:

According to the problem, we have

The combined mass of A and B is 60kg

A's speed is 2m/s

B's speed is 1m/s

The mass of the bag is 5kg

Typically, the momentum of astronaut A along with the bag is defined by

M_A=(60+5)*2

M_A=130kgm/s

To prevent a collision, astronaut A should maintain a speed that is either equal to or less than astronaut B's speed

Thus, the minimum speed astronaut A should achieve corresponds to that of astronaut B, which is 1

Consequently,

130=(60*1)=(5*v)

V=14m/s

7 0
1 month ago
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