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harina [27]
3 years ago
11

Fog rolls in and bright headlights make it hard to see in the fog

Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:F., D.

Explanation:If you use low beams it should help and if you use the defogger it should also help.

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An inductance L, resistance R, and ideal battery of emf are wired in series. A switch in the circuit is closed at time t = 0, at
Kay [80]

Explanation:

After some time t the current does not passing through the circuit

=>so the back emf is zero

=>here the inductor opposes decay of the circuit

- Ldi/dt = Ri

di/dt = - R/Li

di/i = - R/Ldt

now we applying the integration on both sides

log i=-R/Lt+C

here t=0=>i=io

Log io=C

=>Log i=-R/L*t + Log io

logi-Log io=-R/L*t

Log[i/io]=-R/L*t

i/io=e^-Rt/L

i=ioe^-Rt/L

the option D is correct

3 0
3 years ago
Read 2 more answers
The mass of an object is 60kg on the surface of the earth what will be its weight on the surface of the moon
iris [78.8K]

Answer:

Wm = 97.2 [N]

Explanation:

We must make it clear that mass and weight are two different terms, the mass is always preserved that is to say this will never vary regardless of the location of the object. While weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 10 [m/s²]

But in order to calculate the weight of the body on the moon, we must know the gravitational acceleration of the moon. Performing a search of this value on the internet, we find that the moon's gravity is.

gm = 1.62 [m/s²]

Wm = 60*1.62

Wm = 97.2 [N]

8 0
3 years ago
Es frecuente que en las instalaciones eléctricas domésticas se utilice alambre de cobre de 2.05 mm de diámetro. Determine la res
Tasya [4]

Answer:

The resistance is 0.124 ohm.

Explanation:

It is common for domestic electrical installations to use copper wire with a diameter of 2.05 mm. Determine the resistance of such a wire with a length of 24.0 m.

diameter, d = 2.05 mm

radius, r = 1.025 mm

Length, L = 24 m

resistivity of copper = 1.7 x 10^-8 ohm m

Let the resistance is R.

R =\rho \frac{L}{A}\\\\R = \frac {1.7\times10^{-8}\times 24}{3.14\times1.025\times1.025\times 10^{-6}}\\\\R = 0.124 ohm

7 0
3 years ago
A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft
vivado [14]

Their velocity afterwards is 2.88 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

where: in this case:

m_1 = 91.5 kg is the mass of the first player

u_1 = 2.73 m/s is the initial velocity of the first player (choosing east as positive direction)

m_2 = 63.5 kg is the mass of the second player

u_2 = 3.09 m/s is the initial velocity of the second player

v is their combined velocity afterwards

Solving for v, we find:

v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88 m/s

And the sign is positive, so the direction is east.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

7 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
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