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Oxana [17]
2 years ago
11

Compare the density, weight, mass, and volume of a pound of gold to a pound of iron on the surface of Earth.

Physics
1 answer:
Nitella [24]2 years ago
8 0

Answer:

Mass of gold = mass of iron

weight of gold = weight of iron

volume of gold is less than the volume of iron

Explanation:

mass of gold = 1 pound

mass of iron = 1 pound

Mass of both the metals is same.

Weight is equal to the product of mass and the acceleration due to gravity. As acceleration due to gravity is same for both the metals so the weight of both the metals is same.

As the density of iron is less than the density of gold and we know that the volume is defined as the ratio of mass to the density.

mass is same for both the metals, so the volume of iron is more than the density of gold.

Mass of gold = mass of iron

weight of gold = weight of iron

volume of gold is less than the volume of iron

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A climatograph for a tropical grassland or savanna would look different from the climatograph shown for a temperate grassland. D
Savatey [412]

Answer:

Savannas have a fairly constant temperature all year; temperate grasslands have a greater seasonal temperature variation.

Explanation:

For example, the African Savanna has an almost constant temperature all year (see the first figure below).

The difference between summer and winter temperatures is only about 5 °C, and the rate of temperature change is quite slow.

The temperature of a temperate grassland (see the second figure below) has a much greater seasonal variation.  

The summers are hot, and the winters are cold. The difference between summer and winter temperatures is about 30 °C, with a rapid rate of temperature change from one season to the next.

5 0
2 years ago
A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro
MArishka [77]

Answer:

Resistance = 3.35*10^{-4} Ω

Explanation:

Since resistance R = ρ\frac{L}{A}

whereas \rho(x) = a + bx^2

resistivity is given for two ends. At the left end resistivity is 2.25* 10^{-8} whereas x at the left end will be 0 as distance is zero. Thus

2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a

At the right end x will be equal to the length of the rod, so x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25}  = b\\b = 2.77 *10^{-8}

Thus resistance will be R = ρ\frac{L}{A}

where A = π r^2

so,

R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}

6 0
2 years ago
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12
den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s.  What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

0.0192 kgm^{2}/s

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

L = \frac{1}{{12}}m{l^2}\omega

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for \omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}

5 0
2 years ago
Read 2 more answers
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

i_1+i_3=i_2                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

i_3=i_2-i_1=0.75A-0.40A=0.35A

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0
2 years ago
Water is to be boiled at sea level (1 atm pressure) in a 30-cm-diameter stainless steel pan placed on top of a 18-kW electric bu
Tamiku [17]

Answer:

Explanation:

18 kW = 18000 J /s

60% of 18kW = 10800 J/s

Latent heat of evaporation of water

= 2260 x 10³ J / kg

kg of water being evaporated per second

= 10800 / 2260 x 10³ kg /s

= 4.7787 x 10⁻³ kg / s

= 4.78 gm / s .

3 0
2 years ago
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