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zhenek [66]
3 years ago
15

Physical Science Lesson 4 unit 2 questions, I need the answers

Physics
2 answers:
neonofarm [45]3 years ago
7 0
I know I'm a bit late but just in case you still need some of the answers.

1: Oxygen
2: Change in shape
3: Water breaking down into hydrogen and oxygen
4: I think it's Milk but I might be wrong.

I might be wrong about some of them, sorry it's not much.

lubasha [3.4K]3 years ago
5 0

1- Oxygen

2- Change in shape

3- Water breaking down into hydrogen and oxygen

4- Milk

5-Liquid

6- Solid

7-Physical

8- Chemical property

9- Volume

10- Kinetic Energy

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Sarai strolls to the right with an average speed of 1.2m/s for 1500 seconds. What was Sarai's displacement in meters?
Sladkaya [172]

Answer:

1800 meters

Explanation:

Average speed=1.2 m/s

time= 1500 secs

Displacement = average speed × time

= 1.2 × 1500

=1800 meters

5 0
3 years ago
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The three different states of matter are liquid, solid, and gas. A solid is something you can hold in your fingers, while a liqu
nika2105 [10]

The answer is C. that liquids and gases both take the shape of their container.  

Think of it this way, if you take an ice cube and put it in your glass, it will stay in its shape and stay that way until it melts.  But if you put liquid or a gas into a glass, it will take the shape of the glass that it is put into.  

3 0
3 years ago
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Hat processes lead to glacial erosion ?
azamat
Yes......................
8 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by
lana66690 [7]

Answer:

Angle: 48.19^o

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

v_b\ cos\alpha

where v_b is the speed of the boy in still water and \alpha is the angle respect to the shoreline. If the river flows at speed v_s, we now set

v_b\ cos\alpha=v_s

\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}

\alpha=48.19^o

8 0
3 years ago
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