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professor190 [17]
3 years ago
7

Please help!!!!! :)

Physics
2 answers:
kow [346]3 years ago
5 0
The answer is A because bases are greater than 7 and acids are lower than 7
Jlenok [28]3 years ago
4 0
What's up ? ! ?  You should know most of these.

RbOH is a base because its PH is greater than 7.

Sulfur trioxide is SO₃ .

Mixtures B, C, and D can be separated by distillation.
Pebbles and sand can't.

If some salt is dissolved in a beaker of water, then
the salt is the solute and the water is the solvent.

The equation is not balanced because the number of atoms
of each element on the left is not equal to the number of atoms
of that element on the right. 
In fact, the total number of all atoms on the left (8) is not even
equal to the total number of all atoms on the right (10).
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How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)​
frez [133]

Answer:

<h2>98 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 5 × 9.8 × 2

We have the final answer as

<h3>98 J</h3>

Hope this helps you

5 0
3 years ago
Please help! The image produced by a concave mirror is ? .
Alexeev081 [22]

Answer:

is a reflection.

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted). When the object is that the focal point, the image is at infinity.

Explanation:

6 0
3 years ago
Read 2 more answers
An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a
blsea [12.9K]

Answer: apparent weighlessness.


Explanation:


1) Balance of forces on a person falling:


i) To answer this question we will deal with the assumption of non-drag force (abscence of air).


ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).


iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).


2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.


Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0
3 years ago
Read 2 more answers
A stationary boat bobs up and down with a period of 2.4 s when it encounters the waves from a moving boat.
lapo4ka [179]
It seems like the question is asking for the frequency.
Given: 
Time period (T) = 2.4 sec

Frequency (f) =? 

We know that the formula for frequency is:

Frequency (f) = 1/time period (T) 

= 1 / 2.4 s 

= 0.42 Hz. is the frequency for this problem.
8 0
3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
4 years ago
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