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3 years ago

A metre rule is used to measure a length. Which reading is shown to the nearest millimetre? A 0.7m B 0.76m C 0.761m D 0.7614m

Physics

2 answers:

Cloud [144]3 years ago

4 0

Answer:

none

Explanation:

~both of them show to the nearest metre.

~millimeter has (mm) unit eg 0.7mm

Ratling [72]3 years ago

3 0

0.76cm is the correct answer

bcoz after Mili metre there is centi metre so option c is correct

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Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen

mash [69]

Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
$E=hf$
where
h is the Planck constant
f is the photon frequency
Substituting $f=8.88 \cdot 10^{14}Hz$ , we find
$E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J$

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
$\Delta E= -5.86 \cdot 10^{-19} J$

6 0

3 years ago

Read 2 more answers

The quadriceps muscles pull on the patella simultaneously. Below are the forces from each

Nostrana [21]

Based on the calculation of the resultant of vector forces:

the resultant force due to the quadriceps is 1795 N
the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.

<h3>What is the resultant force due to the quadriceps?</h3>

The resultant of more than two vector forces is given by:

F = √Fₓ² + Fₙ²

where:

Fₓ is the sum of the horizontal components of the forces
Fₙ is the sum of the vertical components of the forces
Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
F₁ = 680N, θ = 90 = 30 = 120°
F₂ = 220 N, θ = 90 + 16 = 106°
F₃ = 600 N, θ = 90 + 15 = 105°
F₄ = 480 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55

Fx = -280.6 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55

Fₙ = 1773.1 N

then:

F = √(-280.6)² + ( 1773.1)²

F = 1795.16 N

F ≈ 1795 N

Therefore, the resultant force due to the quadriceps is 1795 N

<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>

From the new information provided:

F₁ = 680N, θ = 90 = 30 = 120°
F₂ = 220 N, θ = 90 + 16 = 106°
F₃ = 600 N, θ = 90 + 15 = 105°
F₄ = 720 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55

Fx = -142.95 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55

Fₙ = 1969.72 N

then:

F = √(-142.95)² + ( 1969.72)²

F = 1974.9 N

F ≈ 1975 N

Therefore, the resultant force due to the quadriceps is 1975 N.

Training and strengthening the vastus medialis results in a greater force of muscle contraction.

Learn more about resultant of forces at: brainly.com/question/25239010

3 0

2 years ago

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

mario62 [17]

Answer:

a . 0.35cm

b. 11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

$\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm$

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

$\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm$

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0

3 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

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3 years ago

Which are the three types of mutations?

WITCHER [35]

Replication, Multiplication, and Substitution.

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