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Brilliant_brown [7]

3 years ago

8

If a truck driver is driving at 30 mph and she makes a U-turn, then starts driving at 30 mph in the opposite direction, did the

driver's speed or velocity change after changing direction? Why?
PLEASE THE QUIZ IS DUE IN A BIT :(

1 answer:

AlekseyPX3 years ago

6 0

Answer:

i dont mow

Explanation:

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What best describes the angle between a changing electric field and the electromagnetic wave produced by it?

MrRa [10]

The correct ans is 1.

A changing electric field produces a changing magnetic field which in turn produces a changing electric field. This means that the source has created a changing electric and magnetic fields, perpendicular to each other, that travel away from the source. Electric field, magnetic field and the EM wave are perpendicular to each other. This gives electromagnetic waves their transverse nature.

7 0

3 years ago

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the

DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 $m/s^2$ (block), aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got N = mg cos θ,

f = $\mu_s N$

= $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$ .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$ ................(v)

Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

6 0

3 years ago

Work is done on an object only if the force and displacement are __________?

miskamm [114]

It’s either movement or work or that’s what a quizzie said

8 0

3 years ago

What happens when Earth rotates on its axis and how long does it take

faust18 [17]

Answer:

You get Day and Night

It takes 24 hour

8 0

3 years ago

Read 2 more answers

Since force is dp/dt, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiati

QveST [7]

Answer:

8.67×10⁻⁶ N/m²

Explanation:

p = Momentum of a photon

E = Energy of a photon

c = Speed of light

I = Intensity of light

Force = dp/dt

$p=\frac{E}{c}$

$\\\Rightarrow F=\frac{\frac{dE}{dt}}{c}$

As given in question

$F=2\frac{\frac{dE}{dt}}{c}$

Now F/A = Pressure

$P=\frac{2I}{c}\\\Rightarrow P=\frac{2\times 1300}{3\times 10^{8}}\\\Rightarrow P=8.67\times 10^{-6}\ N/m^2$

∴ Magnitude of the pressure on the sail is 8.67×10⁻⁶ N/m²

4 0

4 years ago