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3 years ago

Superman takes 8 seconds to stop a runway train over a distance of 58 m using a power of

Physics

1 answer:

elena-14-01-66 [18.8K]3 years ago

5 0

Answer:

Workdone = 600 Kilojoules

Explanation:

Given the following data:

Time = 8 seconds

Power = 75,000 Watts

Distance = 58 m

To find the work done;

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

Thus, work done is given by the formula;

Workdone = power * time

Workdone = 75000 * 8

Workdone = 600,000 = 600 KJ

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A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

aleksley [76]

Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters.

3 0

2 years ago

How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met

PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

= mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

= 23 x 9.18 x 1

= 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

= 23 x 9.18 x 6

= 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

= 23 x 9.18 x 5.7

= 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

= 2681.47

= 2682 (rounding off)

5 0

2 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.