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3 years ago

13

Can someone please help me with science.

1 answer:

vichka [17]3 years ago

5 0

1 is b because a runs 20 and b rus 102 is c

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What do you need to know to be able to determine how far a projectile travels horizontally?

Effectus [21]

As for me, there are two suitable answers for the question represented above and here is a short explanation why I consider these two to be correct :
D. The horizontal velocity of the projectile and <span>B. The length of time before it lands
</span> $v_x = v*cos(x) [

d_x = v_x(t) + ½at²$ -- this led me to answers! Hope everything is clear! Regards!<span>
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3 years ago

What are the 3 formulas you can use for vertical motion for a projectile?

klio [65]

I hope this can help you ask me if you need help again

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3 years ago

There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat

Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator $=heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec$

(b) Total heat absorbed =140 j/sec $=140\times 3600=504000j/hour$

Let the temperature be $\Delta T$

Heat absorbed per hour =504000 $[tex]=400\times 10^3\times \Delta T$

So $\Delta T=\frac{504000}{400000}=1.26K$

8 0

3 years ago

The sum of kinetic energy and potential energy within a system is called

SashulF [63]

Answer:

mechanical energy.

Explanation:

7 0

3 years ago

Read 2 more answers

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}$

F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

$q_3=4.89\times 10^{-6}\ C$

Hence, this is the required solution.

3 0

3 years ago