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yulyashka [42]
3 years ago
12

Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s) The unbalanced equation for the reduction of silver nitrate by copper is shown here. Wh

at is the coefficient for Cu(NO3)2 in the balanced equation?
Physics
1 answer:
cricket20 [7]3 years ago
3 0

Answer:

1

Explanation:

Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag

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On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance
Tresset [83]
Answer B. 112 m



Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m
4 0
2 years ago
Does the depth of a pot affect how water freezes
Iteru [2.4K]
In my opinion it does. The more water the pot holds, the longer you need to wait for it to freeze. Since there is more water, some parts may not be completely frozen. An experiment you can try is to get an ice cube container and a pot. fill both of them and put them in the freezer for the same amount of time. When you take it out, the ice cubes should be frozen leaving the pot with cold water.
3 0
3 years ago
Read 2 more answers
Lowest frequency of wave
Harlamova29_29 [7]
The answer would be a radio wave
4 0
3 years ago
Read 2 more answers
What is not a raquet sport
uysha [10]

Answer:

what

Explanation:

Racket sports include tennis, badminton, squash or any other sport where you use rackets to hit a ball or shuttlecock to play. They can be played competitively or just for fun and are a great form of physical activity.

7 0
3 years ago
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

Average Force, F=144.3396\,N

  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

F=144.3396\,N

Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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