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3 years ago

A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?

Physics

1 answer:

zloy xaker [14]3 years ago

8 0

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

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A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

5 0

3 years ago

To make a given sound seem twice as loud, how should a musician change the intensity of the sound?

Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

$\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}$

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

$I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1$

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0

3 years ago

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th

jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

R = Vo² sin (2θ) / g

sin 2θ = g R / Vo²

sin 2θ = 9.8 75/35²

2θ = sin⁻¹ (0.6)

θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

X = Vox t

t2 = X2 / Vox = X2 / (Vo cosθ)

t2 = 37.5 / (35 cos 18.4)

t2 = 1.13 s

With this time we calculate the height at this point

Y = Voy t - ½ g t²

Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²

Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0

3 years ago

Read 2 more answers

Which of the following statements is true? A. All metals are magnetic. B. All the magnets are made of iron. C. Earth is a giant

harina [27]

C. Earth <em>is </em>a giant magnet

3 0

3 years ago

The unit for energy is the

nikitadnepr [17]

Explanation:

Joule (J) is the MKS unit of energy, equal to the force of one Newton acting through one meter.

4 0

4 years ago