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Vlad1618 [11]
3 years ago
10

calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm​

Physics
1 answer:
Ghella [55]3 years ago
8 0

Answer:

R = 0.21 Ω

Explanation:

the formula:

R = r x l/A

R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)

R = 6.6 x 10-⁷ / 3.14 x 10-⁶

R = 0.21 Ω

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A diffraction grating is to be used to find the wavelength of the emission spectrum of a gas. The grating spacing is not known,
Bumek [7]

Answer:

528.9 nm

Explanation:

For a grating dsinθ = mλ where m = order of grating, d = grating space, λ = wavelength of light and θ = angle of deflection of light

First, we find the grating space d = mλ/sinθ where m = 2 for second order, λ = 632.8 nm = 632.8 × 10⁻⁹ m, θ = 43.2°

d = mλ/sinθ = 2 × 632.8 × 10⁻⁹ m ÷ sin43.2° = 1.849 × 10⁻⁶ m = 1.849 μm

We now find the wavelength of the light to be measured from λ = dsinθ/m

Here, θ = 34.9° and m = 2 for second order. So, we have

λ = dsinθ/m = 1.849 × 10⁻⁶ m × sin34.9° ÷ 2 = 0.5289 × 10⁻⁶ m = 528.9 nm

7 0
3 years ago
give an example of how the law of inertia is demonstrated (a) for moving objects and (b) for objects at rest
lakkis [162]

Answer:

Both

Explanation:

Both of them are part of newtons laws of motion

6 0
2 years ago
During an ice show a 70.0 kg skater leaps into the air and is caught by an initially stationary 55.0 kg skater. What is their fi
mezya [45]

Answer:2.24 m/s

Explanation:

Let the mass of ice skater who is moving is m_1=70kg

mass of stationary skater, m_2=55 kg

conserving momentum

m_1v+0=\left ( m_1+m_2\right )v_f

v_f=\frac{70\times 4}{125}

v_f=\frac{280}{125}=2.24 m/s

5 0
4 years ago
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
lbvjy [14]

Answer:

a) 25 m/s

b) 32 m/s

Explanation:

note:

<u>solution is provided in word attach file please find the attach documents.</u>

Download docx
4 0
4 years ago
A 2200 kg car moving at 31 m/s comes to a complete stop. What is the work done by the breaks of the car?
Free_Kalibri [48]

Answer:

The brakes of this car did approximately \left(-1.1 \times 10^{6}\; \rm J\right) of work. (Rounded to two significant figures. Assumption: this car is moving horizontally, and that all the lost kinetic energy could be attributed to the braking force.)

Explanation:

<h3>Is the work of this braking force positive, zero, or negative?</h3>

The first step is to find out whether the work of the brakes on the car is greater than zero, smaller than zero, or equal to zero. That could be achieved by comparing:

  • the direction of this braking force, and
  • the direction in which the target of the braking force (i.e., the rest of the car) has moved.

In this example, the braking force constantly points backwards. On the other hand, the car kept moving forward (until it stopped.)

In other words, the braking force continuously points in the opposite direction of the motion of its target (the car) until the car came to a stop. Therefore, the work of that this braking force has done should be smaller than zero (i.e., negative.)

<h3>What is the magnitude of the work that this braking force has done?</h3>

Assume that this car is moving horizontally on level ground. Ideally, the amount of work that this braking force has done would be equal to the amount of kinetic energy that this car has lost.

Initial kinetic energy of this car:

\begin{aligned}& \text{KE(initial)}\\ &= \frac{1}{2} \, m \cdot {\left(v(\text{initial})\right)}^{2}\\ &= \frac{1}{2}\times 2200\; \rm kg \times {\left(31\; \rm m \cdot s^{-1}\right)}^{2} \\ &= 1.0571 \times 10^{6}\; \rm J\end{aligned}.

Kinetic energy of this car after it came to a complete stop:

\begin{aligned}\text{KE(final)}= 0\; \rm J\end{aligned}.

Kinetic energy that this car has lost: 1.0571\times 10^{6}\; \rm J.

That should be the same as the magnitude of the work that this braking force has done. However, because this work is smaller than zero, the actual value of this work would be \left(-1.0571\times 10^{6}\; \rm J\right).

In this question, both the mass and the initial speed of this car are specified with only two significant figures each. Hence, round the result accordingly and keep only two significant figures: \left(-1.1\times 10^{6}\; \rm J\right).

7 0
3 years ago
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