Ask question

Ask question

All categories

3 years ago

9

Object C has a mass of 3,600 kilograms. Object D has a mass of 900 kilograms. Both objects were placed on different planets so t

hat their weights are equal.
The gravity acting on Object C is ________ times the gravity acting on Object D.

1⁄6
1⁄4
1
4

1 answer:

FinnZ [79.3K]3 years ago

7 0

The correct answer would be 4

You might be interested in

Which is the BEST example of the kind of mechanics that are studied in sports biometrics?

ikadub [295]

Answer:

-A.

Explanation:

: Hope it's Help:

[correct me if I'm not correct]

5 0

1 year ago

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>

6 0

3 years ago

A wire is made from a material having a temperature coefficient of resistivity of 0.0003125 (°C^-1). In an experiment, we mainta

beks73 [17]

Answer:

The power decreases by 36%

Explanation:

Given:

At 20° C

Power, P₀ = 300 W

Potential difference, V = 150 volts

Now, power is given as

P = V²/R

where, R is the resistance

on substituting the values, we get

300 = 150²/R₀

or

R₀ = 75 Ω

Now, the variation of resistance with temperature is given as

R = R₀[1 + α(T - T₀)]

where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)

now, at

T₀ = 20° C

R₀ = 75 Ω

for

T = 1820° C

we have

R = R₀[1 + α(T - T₀)]

substituting the values

we get

R = 75×[1 + 0.0003125 × (1820 - 20)]

or

R = 117.18 Ω

Now using the formula for power

We have,

P = V²/R

or

P = 150²/117.18 = 192 W

Therefore, the percentage change will be

= $\frac{P-P_0}{P_0}\times 100$

on substituting the values , we get

= $\frac{192-300}{300}\times 100$

= -36%

here, negative sign depicts the decrease in power

3 0

2 years ago

Read 2 more answers

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago

Please answer the one you know!

tresset_1 [31]

#8 positive kinetic energy

3 0

2 years ago