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3 years ago

An electromagnetic wave moves through electric and magnetic fields. How might the movement of the wave be described?

Physics

1 answer:

Elenna [48]3 years ago

7 0

A. The vibrations of the fields are perpendicular to the direction in which the wave moves.

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On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

3 years ago

Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d

Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

$h_{fg}$ = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

$T_{L}$ = -18.77°C = 254.23 K

At saturation pressure 800 kPa, the saturation temperature is

$T_{H}$ = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, $q_{reject}$ = $h_{fg}$ = 171.82 kJ/kg

We know COP of heat pump

COP = $\frac{T_{H}}{T_{H}-T_{L}}$

= $\frac{304.31}{304.31-254.23}$

= 6.076

Therefore, Work out put, W = $\frac{q_{reject}}{COP}$

= 171.82 / 6.076

= 28.27 kJ/kg

8 0

3 years ago

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

3 0

3 years ago

A person with mass 65 kg, standing still, throws an object at 4 m/s. If the

fenix001 [56]

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(65 kg) (0 m/s) + m (0 m/s) = (65 kg) (-3.5 m/s) + m (4 m/s)

m ≈ 57 kg

7 0

3 years ago

A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the

Maksim231197 [3]

25 kg and 299N are the answers

8 0

3 years ago

Read 2 more answers