All categories

3 years ago

There is up to 30 times more gold in a tons of old mobile phones than in a tons of gold ore. true or false

Physics

1 answer:

FrozenT [24]3 years ago

3 0

Answer:

true

Explanation:

dbgkdodocofkci ifkdcl k kfododocp v

You might be interested in

Two trains, each having a speed of 33 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h fli

Alex Ar [27]

Answer:

66 km

Explanation:

Given that:

The speed of the two trains = 33 km/h

The speed of the bird = 60 km/h

The distance apart between the two trains = 60 km

From the given information, we are being told that the two trains are going at the same speed. Therefore, they will definitely collide at 30 km

We know that:

speed of the train = distance traveled × time

Making the time t the subject of the formula:

time = speed of the train / distance traveled

time = 30 km / 33 km/h

time = 0.909 / hr

Thus, the bird flying at a given speed of 60 km/h in a time of 0.909 / hr will cover a total distance of :

distance (d) = speed of the bird/ time

distance (d) = $\dfrac{60 \ km/hr}{0.909 \ /hr}$

distance (d) = 66 km

3 0

3 years ago

A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.6

dsp73

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

$\sum F = ma$

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

$F - F_f = ma$

To find the required force then,

$F=F_f+ma$

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

$F = \mu mg +ma$

$F = 0.35*50*0.8+50*1.2$

$F=(171.5N)+(50Kg)(1.2m/s^2)$

$F=231.5N$

$F\approx 230N$

Therefore the horizontal force applied on the block is B) 230N

6 0

4 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

7 0

3 years ago

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

4 years ago

What is the frequency of highly energetic ul-

Mnenie [13.5K]

Frequency = (speed) / (wavelength)

Frequency = (3 x 10⁸ m/s) / (124 x 10⁻⁹ m)

Frequency = <em>2.42 x 10¹⁵ Hz</em>

8 0

3 years ago