3 years ago

CO + 7H2 - C3Hg + 3H20

Physics

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

Espanol 7h2 that's c coupon

You might be interested in

We reduce friction in machines? why<br>

kvv77 [185]

Answer:

friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.

5 0

3 years ago

Read 2 more answers

Gravity is a force that pulls things

Katena32 [7]

Answer:

down and towards it

Explanation:

just because it is man

8 0

3 years ago

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

3 years ago

Jessie creates a model of two waves that have the same amplitude as shown.

ankoles [38]

Answer:

Explanation:

7 0

3 years ago

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33×10−21 n

qwelly [4]

Answer:

The number of excess electrons on each sphere is 759

Explanation:

Given that,

distance , d = 20 cm

= 0.20 m

let the number of electrons is n

Electric force (F) = k × (n × e)² /d²

3.33 × $10^{-21}$ = 9 × $10^{9}$ × (n × 1.602 × $10^{-19}$ )² /0.2²

solving for n

n = 759

4 0

4 years ago