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1 year ago

Define couple and give 2 examples

2 answers:

6 0

A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. ... For example, the forces that two hands apply to turn a steering wheel are often (or should be) a couple. Each hand grips the wheel at points on opposite sides of the shaft.

Elodia [21]1 year ago

3 0

Answer:

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation.

For example, turning a key in a lock and turning a steering wheel.

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10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago

Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s

vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances

4 0

2 years ago

A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are

Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0

2 years ago

A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer

Mekhanik [1.2K]

The initial volume of the gas is
$V_i = 600 cm^3$
while its final volume is
$V_f = 200 cm^3$
so its variation of volume is
$\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3$

The pressure is constant, and it is
$p=400 kPa = 400 \cdot 10^3 Pa$

Therefore the work done by the gas is
$W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J$
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
$Q=+100 J$

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
$\Delta U = Q-W = 100 J - (-160 J)=+260 J$
where the positive sign means the internal energy of the gas has increased.

7 0

2 years ago

550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c

Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work

7 0

2 years ago