All categories

2 years ago

(b) The density of aluminum is 2.70 g/cm3. The thickness of a rectangular sheet of aluminum foil varies

Physics

1 answer:

vekshin12 years ago

6 0

Answer:

(i) $22.48 cm^3$

(ii) $1.5 mm$

Explanation:

Let t be the average thickness of the sheet.

Given that:

Density of the aluminum sheet is $2.70 g/cm^3$

Mass of sheet = 60.7 g

Length of sheet = 50.0 cm

Width of sheet = 30.0 cm

(i) Using, Density=Mass/Volume

$\Rightarrow \text{Volume}=\frac{\text{Mass}}{\text{Density}}$

$\Rightarrow \text{Volume}=\frac{60.7}{2.7}=22.48 cm^3$

Hence, the volume of the sheet is $22.48 cm^3$ .

(ii) Now, as this aluminum sheet is in the shape of a cuboid, so the volume of the sheet is

$\text{Volume}=\text{Length}\times\text{Width}\times\text{Thickness}$

$\Rightarrow 22.48=50\times 30 \times w$

$\Rightarrow w=\frac{22.48}{50\times 30}=0.015 cm$

Hence, the average thickness of the sheet is $1.5 mm$ .

You might be interested in

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

2 years ago

For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen

Talja [164]

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

5 0

2 years ago

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

mina [271]

Answer:

$\frac{dQ}{dt}= 4312 W$

Explanation:

As we know that base of the slab is given as

$A = 11 \times 8$

$A = 88 m^2$

now we know that rate of heat transfer is given as

$\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)$

here we know that

$k = 1.4 W/m k$

Also we have

$x =0.20$

$\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)$

$\frac{dQ}{dt}= 4312 W$

7 0

2 years ago

slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of

Ivan

Answer: 800N

Explanation:

Given :

Mass of ball =0.8kg

Contact time = 0.05 sec

Final velocity = initial velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;

Force(F) = mass(m) * average acceleration(a)

a= (initial velocity(u) + final velocity(v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Therefore,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Therefore,

Magnitude of average force (F)

F=ma

m = mass of ball = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

6 0

2 years ago

a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the

Katena32 [7]

Answer:

$W_{earth} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 5.91 * 10^6 N$ : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

$m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg$

$g_{moon}= 1.62 \frac{m}{s^2}$

$g_{earth}= 9.8 \frac{m}{s^2}$

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

$W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N$ : Rocket weight on moon

7 0

2 years ago