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3 years ago

Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?

1 answer:

8 0

Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

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A space shuttle takes off from FL and circles Earth several times, finally landing in CA. While the shuttle is in flight, a phot

mixer [17]

Answer:

Both the astronauts and photographer have the same displacement

Explanation:

Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.

Hence, both the astronauts and photographer will have the same displacement.

3 0

3 years ago

An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan

DanielleElmas [232]

Answer:

0.872m/s

Explanation:

Tangential velocity is given by the formula,

$v= 2\pi r/ t$

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

$\frac{1lap* 2 *3.14 *25m}{180secs}$

v = 157m/180secs

Therefore, the magnitude of the tangential velocity

=0.872m/secs

5 0

2 years ago

A student places his hand in front of a plane mirror as shown in the diagram. Which terms correctly describe the image that the

ANTONII [103]

Either A or D but maybe D

4 0

2 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.