Ask question

Ask question

All categories

3 years 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?

2 answers:

Verdich [7]3 years ago

7 0

Answer:

Explanation:

If we pull the rope by a force F1 then the tension in the rope is same as that of F1.

T1 = F1 .

This forms an action reaction pair.

mixer [17]3 years ago

5 0

Answer:

F.

Explanation:

Here in the question the mass of the pulley is zero, hence, the tension in the cable throughout is same.

magnitude of tension in rope 1 is

T1= F

Hence the tension T1 is rope 1 is F.

You might be interested in

Please help it’s easy will give brainlist !! Thank youu worth a lot be serious pls

Genrish500 [490]

Answer:

Part 1

Stationary

Part 2

20 Newtons

Part 3

Force

Part 4

4.0 m/s²

Part 5

Normal

Part 6

Cart m

Part 7

The gravitational force is less than magnetic force

Explanation:

Part 1

The position time graph of the object is an horizontal straight line passing across the top of the position 3 boxes vertically up from the origin

As the time increases by the units of number boxes to the left, the position of the object does not change and remains at the 3 boxes up above the origin, therefore, the object is stationary

Part 2

By Newton's third law of motion, the action action obtained from a force is equal to the reaction given to the force, therefore, we have;

The force exerted by the student on the scale = The force exerted by the scale on the student = 20 N

Part 3

A force is a the directional push on an object or pull from the object as a form of interaction with another object which tends to alter or maintain the motion of the object

Part 4

The given parameters are'

The mass of block A = 1.0 kg

The mass of block B = 2.0 kg

Both blocks, "A" and "B" are initially at rest

The applied horizontal force, F = 12-N

The nature of the surface over which the blocks move = Smooth surface

Force, F = Mass, m × Acceleration, a

F = m × a

The blocks two blocks experience a common acceleration, a

The combined mass of the two blocks, m = 1.0 kg + 2.0 kg = 3.0 kg

m = 3.0 kg

Therefore, a = F/m = 12-N/(3.0 kg) = 4 m/s²

Part 5

A normal force is a force acting perpendicularly to a surface that supports the weight of an object

Part 6

The given parameters are;

M = 2 kg

The mass of the left cart attached to the spring = M = 2 kg

The mass of the right cart attached to the spring = 2·M = 2 × 2 kg = 4 kg

Therefore, given that the force exerted by one cart on the other after the spring is removed, we have;

Force, F = M × a₁ = 2·M × a₂

Where;

a₁ = The average acceleration of the cart with mass, M

a₂ = The average acceleration of the cart with mass, 2·M

M × a₁ = 2·M × a₂

∴ a₁ = 2·a₂

The acceleration, a₁, of car M = 2 × The acceleration, a₂, of car 2·M

The acceleration of cart M is two times the acceleration of cart 2·M

Cart M will experience a greater average acceleration

Part 7

For the top magnet of two magnets placed two magnets arranged so that one is on top the other, to be held in the air with a space between the two magnets, the gravitational force pulling the top magnet down, is less than the repelling magnetic force between the two magnets.

7 0

3 years ago

Pablo and Charles were conducting an investigation where they were measuring the energy of a glass marble as it rolled down a ra

Fantom [35]

Answer:

it will be B

Explanation:

3 0

3 years ago

Read 2 more answers

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

3 years ago

An Ohmic material is one in which its resistance does not depend on the voltage across it or the current within it. In that case

Alexandra [31]

Voltage

Mark brainliest

6 0

4 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

3 0

3 years ago