Sunglasses that reduce glare take advantage of which kind of wave

1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0

igor_vitrenko [27]

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0 m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

$\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2$

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m$

So, the plane slide a distance of 229.5 m.

6 0

3 years ago

Two objects are attracted to each other by electromagnetic forces. Suppose I double the charge of one object. How can I return t

nata0808 [166]

Answer:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

Explanation:

Lets assume,

Charge on first object = Q

Charge on second object = q

Distance between them = r

Force between the two charges = F

According to Coulomb's law,

$F = k \frac{Qq}{r^{2}}$

where, k = Coulomb constant

New value of charge on first object = 2Q. Thus the new force(F') will be

$F' = k \frac{2Qq}{r^{2}}$

$F' = 2F$

So, to bring the value of force(F') to original value, there are two options:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

4 0

3 years ago

Two long, straight wires are separated by a distance of 32.2 cm. One wire carries a current of 2.75 A, the other carries a curre

Igoryamba

Answer:

a) $\frac{F_1}{L}=1.95*10^-^5N$

b) $\frac{F_2}{L}=1.95*10^-^5N$

Explanation:

From the question we are told that:

Distance between wires $d=32.2$

Wire 1 current $I_1=2.75$

Wire 2 current $I_2=4.33$

a)

Generally the equation for Force on $l_1$ due to $I_2$ is mathematically given by

$F_1=I_1B_2L$

Where

B_2=Magnetic field current by $I_2$

$B_2=\frac{\mu *i_2}{2\pi d}$

Therefore

$F_1=I_1B_2L$

$F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L$

$\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }$

$\frac{F_1}{L}=1.95*10^-^5N$

b)

Generally the equation for Force on $I_2$ due to $I_1$ is mathematically given by

$F_2=I_2B_1L$

Where

B_1=Magnetic field current by $I_2$

$B_1=\frac{\mu *I_1}{2\pi d}$

Therefore

$\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})$

$\frac{F_2}{L}=1.95*10^-^5N$

5 0

3 years ago

Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the

Lelechka [254]

Answer:

Ft = 17.48°C

Explanation:

Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

Therefore,

Heat absorbed by water of 20g = heat rejected by water of 265g.

So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)

So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)

To get;

7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

So factorizing, we get;

83.74 Ft + 1193.3 Ft = 29832.4 - 7513

So; 1277.04 Ft = 22319.4

So; Ft = 22319.4/1277.04 = 17.48°C

7 0

3 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

Kipish [7]

Answer:

The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

6 0

3 years ago