Which statements represent properties of intermolecular forces? Select all that apply.

Nerf this hbhbhbbhbbhbhbhbhbhbhb

Vanyuwa [196]

Answer:

.......................

Explanation:

what

6 0

3 years ago

Read 2 more answers

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

dangina [55]

1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.

2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
$f'= \frac{c}{c+v_s} f_0$
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.

6 0

3 years ago

A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B ) , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

= - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0

3 years ago

A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13

Irina18 [472]

The work done by the gravitational force = 0

Given the mass of the box = 40 kg

The box is initially at rest.

Distance moved by the applied force = 5m

Force applied = 130 N

Co-efficient of friction between the box and floor = 0.3

The box is moved only in the horizontal direction by the applied force.

Gravitational force is applied in a direction perpendicular to the applied force. hence it doesn't do any work on the box.

Therefore, the work done by the gravitational force is zero.

Learn more about the gravitational force at brainly.com/question/862529

#SPJ4

3 0

2 years ago

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

Alex

Answer:

616.223684211 N

Explanation:

$F_r$ = Resistive force on the wheel = 115 N

F = Force acting on sprocket

$r_2$ = Radius of sprocket = 4.75 cm

$r_1$ = Radius of wheel = 25 cm

Moment of inertia is given by

$I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2$

Torque

$\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm$

Torque is given by

$\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N$

The force on the chain is 616.223684211 N

5 0

3 years ago