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skelet666 [1.2K]
1 year ago
10

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

ht line. For this situation, which of the following statements are true? Check all that apply. 1. The tension in the rope is everywhere the same. 2. The magnitudes of the forces exerted on the two objects by the rope are the same.3. The forces exerted on the two objects by the rope must be in opposite directions.4. The forces exerted on the two objects by the rope must be in the direction of the rope.
Physics
1 answer:
Virty [35]1 year ago
6 0

Answer:

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a<em> "net force"</em> which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. <u>So this explains number 2 & 3 answers.</u>

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). <u>So, this explains number 1.</u>

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Which condition in a nebula would prevent nuclear fusion?
Tanzania [10]
<span>A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.</span>
8 0
2 years ago
Read 2 more answers
The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan
Karo-lina-s [1.5K]

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

8 0
2 years ago
A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going
tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2

For the sound

Distance = Speed × Time

\text{Distance}=340\times (2.5-t)

Here, distance traveled by the stone and sound is equal

\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0

t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds

The time taken to fall down is 2.4158 seconds

h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m

Height of the buidling is 28.6260196842 m

7 0
2 years ago
The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr
VladimirAG [237]

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

<u>Explanation:</u>

Given data,

E= 3 ×10 ⁶ Δx=0.06/100

We have to find the minimum potential difference

E= -ΔV/Δx

ΔV=- E × Δx

ΔV =-3 ×10 ⁶ . 0.06/100

ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

6 0
1 year ago
Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to fi
JulijaS [17]

Answer:

  b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

               fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

               w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

             

let's call

               w₀² = k / m

               w² = w₀² - b² / 4m²

               b² = (w₀² -w²) 4 m²

Let's find the angular velocities

             w₀² = 5 / 0.1

             w₀² = 50

             w = 2π f

             w = 2π 1

             w = 6.2832 rad / s

we subtitute

               b² = (50 - 6.2832²) 4 0.1²

               b = √ 0.42086

                b = 0.6487 kg / s

8 0
1 year ago
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