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3 years ago

Tina lifted a box 0.5 m. the box weighed 25 n. how much work did tina do on the box?

Physics

1 answer:

sweet [91]3 years ago

5 0

Work= F times d so force x distance
25(.5) is 12.5

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If something is hot, does it contain heat or thermal energy?

zysi [14]

I am absolutely sure that if something is hot, it contain thermal energy only because <span>if a given body is heated its electrons start to move very fast which means that thermal energy is being produced. Hope everything is clear! Regards!</span>

8 0

3 years ago

Find the distance along an arc on the surface of the earth that subtends a central angle of 1 minutes (1 minute = 1/60 degree).

-Dominant- [34]

Answer:

1.152 miles

Explanation:

Given: central angle = 1 minute = $(\frac{1}{60}) ^{o}$

radius of the earth = 3960 miles

The length of an arc = $\frac{\alpha }{360^{o} }$ 2 $\pi$ r

where: $\alpha$ is the central angle, and r is the radius.

Thus,

Distance along the arc = $\frac{\alpha }{360^{o} }$ 2 $\pi$ r

Distance along the arc = $\frac{(\frac{1}{60}) ^{o} }{360^{o} }$ x 2 x $\frac{22}{7}$ x 3960

= $\frac{(\frac{1}{60}) ^{o} }{360^{o} }$ x 24891.4286

= 1.1524

The required distance along an arc is 1.152 miles.

4 0

4 years ago

A spherical conducting shell has charge Q. A point charge q is placed at the center of the cavity. The charge on the inner surfa

FinnZ [79.3K]

Answer:d

Explanation:

From Gauss law Electric field inside a surface is directly proportional to the charge enclosed in it.

Electric field inside a spherical shell is zero and hence there is no charge inside the spherical shell because q charge induces a -q charge on inside surface of spherical shell.

and to counter it there is q charge on the surface. So total charge outside the surface is Q+q

7 0

3 years ago

-Para lograr que una pieza de 0,300 kg de cierto metal aumente su temperatura desde 40°C a 60 °C ha sido necesario suministrarle

marishachu [46]

Answer:

parte a) el calor específico es c = 0.383 J /(gr*K)

parte b) la temperatura inicial es T inicial= 52.72 °C

Explanation:

para el primer punto la formula para el calor Q es:

Q = m * c * ( T final - T inicial )

donde

m= masa de la pieza = 0.300 kg = 300 gr

Q = flujo de energía en forma de calor= 2299 J

c = calor específico

T final = temperatura final =40°C

T inicial = temperatura inicial = 60 °C

entonces

Q = m * c * ( T final - T inicial )

c = Q / [ m* ( T final - T inicial ) = 2299 J/[ 300 gr * ( 60 °C - 40°C )]

= 0.383 J /(gr*K)

c = 0.383 J /(gr*K)

para el segundo punto usamos la misma formula

Q = m * c * ( T final - T inicial )

pero

m= 200 gr= 0.200 kg

c=459.8 J/(kg*K) , Q =20.900 J , T final = 280 °C

Q = m * c * ( T final - T inicial )

T inicial = T final - Q/(m*c) =280 °C - 20.900 J/(459.8 J/(kg*K)* 0.200 kg) = 52.72 °C

7 0

4 years ago

Whta is the mechanical advantage when the effort force is 20 and the resistance force is 5

pogonyaev

The mechanical advantage would be 4. 20/5=4

7 0

3 years ago