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Free_Kalibri [48]

3 years ago

11

Tom stops using a new skin cream for five days to determine if the cream is the cause of a recent rash. What stage of the scient

ific method is Tom using?

1 answer:

faltersainse [42]3 years ago

5 0

I think this is simply the stage of observation

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Which type of rock would form from a lava following a volcanic eruption

melisa1 [442]

When lava reaches the surface of the Earth through volcanoes or through great fissures the rocks that are formed from the lava cooling and hardening are called extrusive igneous rocks. Some of the more common types of extrusive igneous rocks are lava rocks, cinders, pumice, obsidian , and volcanic ash and dust.

5 0

4 years ago

1. Take the speed at which the body moves as 10 m/s 2. Calculate the distance travelled by the body every second 3. Take the tim

exis [7]

Answer:

a)this graph is also a line b) in both cases we have a uniform movement

Explanation:

In this exercise we have a uniform movement

v = d / t

d = v t

in the table we give some values to make the graph

t (s) d (m)

1 10

2 20

3 30

In the attached we can see the graph that is a straight line

we have another vehicle at v = 50 me / S

t (s) d (m)

1 50

2 100

3 150

this graph is also a line

b) in both cases we have a uniform movement

3 0

4 years ago

Given a 3.00 μF capacitor, a 7.75 μF capacitor, and a 5.00 V battery, find the charge on each capacitor if you connect them in t

USPshnik [31]

Answer:

a) Q1= Q2= 11.75×10^-6Coulombs

b) Q1 =15×10^-6coulombs

Q2 = 38.75×10^-6coulombs

Explanation:

a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as

1/Ct = 1/C1 + 1/C2

Given C1 = 3.00 μF C2 = 7.75μF

1/Ct = 1/3+1/7.73

1/Ct = 0.333+ 0.129

1/Ct = 0.462

Ct = 1/0.462

Ct = 2.35μF

V = 5.00Volts

To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance

Q = 2.35×10^-6× 5

Q = 11.75×10^-6Coulombs

Since same charge flows through a series connected capacitors, therefore Q1= Q2=

11.75×10^-6Coulombs

b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2

C = 3.00 μF + 7.75 μF

C = 10.75 μF

For 3.00 μF capacitance, the charge on it will be Q1 = C1V

Q1 = 3×10^-6 × 5

Q1 =15×10^-6coulombs

For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5

Q2 = 38.75×10^-6coulombs

Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.

7 0

3 years ago

What is the relevant similarity of the following argument?

meriva

B. Tolerating a bully or dictator leads to further abuse

3 0

3 years ago

A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

$\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h$

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

$V=\dfrac{x}{t}$

Where x is the distance, t is the time and v is the speed. using the data that we have:

$V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h$

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

$V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km$

7 0

3 years ago