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2 years ago

Another word for a joule per second is a(n)

Physics

1 answer:

marishachu [46]2 years ago

5 0

The watt (symbol: W) is the SI derived unit of power, equal to one joule of energy per second

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A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:

Igoryamba

Answer:

$9.8\; {\rm J}$ , assuming that the gravitational field strength is $g = 9.8\; {\rm N \cdot kg^{-1}}$ .

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be $0$ . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let $m$ denote the mass of this block. It is given that $m = 1\; {\rm kg}$ . The weight of this block would be:

$\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}$ .

Hence, the force that the boy applies on this block would be upward with a magnitude of $F = 9.8\; {\rm N}$ .

The mechanical work that a force did is equal to the product of:

the magnitude of the force, and
the displacement of the object in the direction of the force.

The displacement of this block (upward by $s = 1\; {\rm m}$ ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:

the magnitude of the force that this boy exerted, $F = 9.8\; {\rm N}$ , and
the displacement of this block in the direction, $s = 1\; {\rm m}$ .

$\begin{aligned}\text{work} &= F\, s \\ &= 9.8\; {\rm N} \times 1\; {\rm m} \\ &= 9.8\; {\rm J}\end{aligned}$ .

5 0

1 year ago

Help and show work thanks

Zepler [3.9K]

Well its C, cant show you the work its in my head sorry.

6 0

2 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

5 0

3 years ago

During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a

lana66690 [7]

According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.

8 0

3 years ago

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh

SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0

2 years ago