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katen-ka-za [31]
2 years ago
11

A long thing bar of copper is heated evenly along it's length

Physics
1 answer:
leva [86]2 years ago
8 0

Answer:

It becomes less heavy

Explanation:

Hopefully it helps!

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Cocking your head would be most useful for detecting the ______ of a sound.
statuscvo [17]
Cocking your head would be most useful for detecting the LOCATION of a sound.
5 0
3 years ago
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A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resistin
otez555 [7]

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

a=3 \ m/s^2

Forces resisting motion,

F_f=432 \ N

Mass,

m = 241 \ kg

By using Newton's second law, we get

⇒ F-F_f=ma

Or,

⇒         F=ma+F_f

By putting the values, we get

⇒             =(3\times 241)+432

⇒             =723+432

⇒             =1155 \ N

7 0
3 years ago
Look at the advertisement. A poster that reads, "There's more to resistors than resistance. If you're really serious about cost,
ozzi

Options:

universal appeal

flattery

association

bandwagon

Answer: Association

Explanation: Advertisement is a marketing technique through which business organisations utilize the opportunities provided by both the print, electronic and other channels of communication to market a product to the target audience.

Association Advertising is a type of Advertising where certain attributes which have been known to be associated with good and quality products are used to market the products to the target audience.

7 0
4 years ago
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The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and
Lunna [17]

Answer:

Q = - 4312 W = - 4.312 KW

Explanation:

The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:

Q = - kA dt/dx

Integrating from one side of the slab to other along the thickness dimension, we get:

Q = - kA(T₂ - T₁)/L

Q = kA(T₁ - T₂)/t

where,

Q = Rate of Heat Loss = ?

k = thermal conductivity = 1.4 W/m.k

A = Surface Area = (11 m)(8 m) = 88 m²

T₁ = Temperature of Bottom Surface = 10°C

T₂ = Temperature of Top Surface = 17° C

t = Thickness of Slab = 0.2 m

Therefore,

Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m

<u>Q = - 4312 W = - 4.312 KW</u>

<u>Here, negative sign shows the loss of heat.</u>

3 0
4 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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