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1 year ago

If a crane worker lifts a crate with a mass of 250 kg from the ground to a

Physics

1 answer:

mina [271]1 year ago

7 0

The gravitational of a body is possessed by the body due to the virtue of its position.

The formula for gravitational potential energy is,

P.E = mgh joules

Substituting the values

196*9.8*250= 49k

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Direct current, DC, flows in a _________ direction while alternating current, AC, the direction of flow ___________. A) single,

kramer

The flow of Direct current (DC) is constant and flows in one direction. Most digital electronics make use of DC. Alternating current (AC) periodically flows in reverse and is mostly used to deliver power to houses, buildings and the like. With that alone, you can already rule out A, C and D.

The answer would then be B. constant, periodically reversing.

4 0

3 years ago

Read 2 more answers

Please indicate how long each bar is in centimeters and millimeters.

Maru [420]

Answer:

1) 74 cm or 740 mm

2) 33 cm or 330 mm

3) 76 cm or 760 mm

4) 17 cm or 170 mm

brainliest !?!?

6 0

2 years ago

What happens to a path of a light ray parallel to the principal axis, after it passes through a converging

Simora [160]

Answer: The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. ... All three rays should intersect at exactly the same point.

Explanation: Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses.

3 0

2 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

2 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

5 0

2 years ago