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3 years ago

How many hydrogen atoms are there in all the reactants?

Physics

2 answers:

Alinara [238K]3 years ago

6 0

Answer:

There are two hydrogen atom in all the reactants.

11Alexandr11 [23.1K]3 years ago

6 0

Answer:

we can find two hydrogen atom in all the reactants

Explanation:

pill!!

You might be interested in

An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r

alexandr402 [8]

Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it $f_e$
and we need the Reflected = $f_{r}$
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz = $f_{b}$
so we have:
$f_{e}$ - $f_{r}$ = $f_{b}$
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = $f_{r}$
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=> $v_{max}$ = (lambda)( $f_{r}$
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the $v_{max}$ is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0

4 years ago

Directions: I answer all the questions listed below.

anzhelika [568]

Answer:

The compression ratio is 10

Explanation:

Given

$V_d = 450cm^3$ --- swept volume

$V_c = 50cm^3$ --- compression volume

Required

Determine the compression ratio (CR)

This is calculated as:

$CR = \frac{V_d + V_c}{V_c}$

$CR = \frac{450+50}{50}$

$CR = \frac{500}{50}$

$CR = 10$

5 0

3 years ago

Read 2 more answers

A plant uses energy from the Sun to make food. What kind of energy transformation is this?

Katarina [22]

Light to chemical (C)

4 0

4 years ago

Read 2 more answers

A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has ra

kaheart [24]

Answer:

The plank moves 0.2m from it's original position

Explanation:

we can do this question from the constraints that ,

the wheel and the axle have the same angular speed or velocity
the speed of the plank is equal to the speed of the axle at the topmost point .

thus ,

since the wheel is pure rolling or not slipping,

⇒ $v=wr$

where

$v$ - speed of the wheel

$w$ - angular speed of the wheel

$r$ - radius of the wheel

since the wheel traverses 1 m let's say in time ' $t$ ' ,

$v_{w}=\frac{distance}{time} =\frac{1}{t}$

∴

⇒ $w=\frac{v_{w}}{r} = \frac{1}{t*0.25}$

the speed at the topmost point of the axle is :

⇒ $v_{a}=w*r\\v_{a}=\frac{1}{t*0.25} *0.05\\v_{a}=\frac{1}{5t}$

this is the speed of the plank too.

thus the distance covered by plank in time ' $t$ ' is ,

⇒ $d=v_{a}*t\\d=\frac{1}{5t} *t\\d=\frac{1}{5} = 0.2m$

8 0

4 years ago

Need help with both questions!

xenn [34]

#14 isn't really a Physics problem. It's more of just reading a graph.

A). When speed changes, acceleration is

(change in speed) / (time for the change) .

To be correct about it, acceleration can be positive ... when speed
is increasing ... or it can be negative ... when speed is decreasing.
So, on this graph, there are two periods of acceleration:

From zero to 2 seconds, acceleration = (8 m/s) / (4 sec) = 2 m/s² .

From 10 to 12 seconds, acceleration = (-4 m/s) / (2 sec) = -2 m/s² .

B). From 12 to16 seconds, you can read the speed right from
the graph. It's 4 m/s .

C). From 2 to 10 seconds, the objects speed is a steady 8 m/s.
Covering 8 m/s every second for 8 seconds, it covers 64 meters.
Do you remember that distance is the area under the speed/time
graph? You can see that plainly on this graph. From 2 to 10 sec,
there are 16 blocks. Each block is (2 m/s) high and (2 sec) wide,
so its area is (2 m/s) x (2 sec) = 4 meters. The area of 16 blocks
is (16) x (4 meters) = 64 meters.
====================================

#15.

a). constant velocity on a distance graph is a line that slopes up;
constant velocity on a velocity graph is a horizontal line;

b). positive constant acceleration on a distance graph is a
line that curves up;
positive constant acceleration on a velocity graph is a
straight line that slopes up;

c). "uniformly slowing down to a stop" on a distance graph
is a line that's less and less curved as time goes on, and
eventually reaches the x-axis.
"uniformly slowing down to a stop" on a velocity graph is
a straight line that slopes down, and stops when it reaches
the x-axis.

7 0

3 years ago