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3 years ago

14

Taylor drives 5 miles in 10 minutes. She stops at a light for 2 minutes. She then travels another 10 miles in 8 minutes. What wa

s her average speed?

1 answer:

attashe74 [19]3 years ago

5 0

45 miles per hour is the answer to the probably.

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Oxygen gas is used to fill a balloon and a short time later the balloon is observed to only be half full. Explain this observati

lesya692 [45]

Answer:

It would be A.

Explanation:

When oxygen is released, that oxygen rises and the ballons temperature will go down. Heat rises so if the ballon is dropping, then the temperature is dropping.

3 0

3 years ago

Read 2 more answers

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

algol13

Answer: 38.2 μC

Explanation: In order to solve this problem we have to use the relationship for a two plate capacitor with a dielectric so:

C= Q/V= we also know that for two paralel plates C=εo*k*A/d and V=E/d

where k is the dielectric constant, A plate area, V is potential difference; E electric field and d the separation between the plates.

reorganizing we have:

Q/A=σ= E*k/εo= 1.2 * 10^6*3.6/8.85 * 10^-12=38.2μC

7 0

4 years ago

Earth's atmosphere contains only small amounts of carbon dioxide because

Mamont248 [21]

Of biogeochemical cycles

Because certain biogeochemical cycles take place to lessen these gases from forming since they are not beneficial for life if too much.
Thank you for your question. Please don't hesitate to ask in Brainly your queries.

4 0

3 years ago

Overload refers to: A. Performing a weight-lifting exercise with the resistance (load) held overhead B. Using a demand (load) ab

andrey2020 [161]

Answer:

E. All of the answers are correct

Explanation:

Overload principle in fitness training is associated with a gradual development of an athlete's abilities by progressively increasing the athlete's load and training.

In order to do this, the athlete's limit must be surpassed albeit gradually at first then picked up later over time.

6 0

3 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