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Aliun [14]
3 years ago
15

According to Newton's third law of motion, if you push against a wall, the wall will __________.

Physics
2 answers:
slavikrds [6]3 years ago
3 0

Answer:

push up agaisnt you with equal force.

egoroff_w [7]3 years ago
3 0

Answer:

c

Explanation:

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What are some errors made while doing measuring with a triple beam balance lab
nydimaria [60]
Make sure the triple beam balance is at 0 before you begin.
7 0
3 years ago
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Calculate the force that the 4kg block exerts on the 10kg block
Kryger [21]
Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.
8 0
3 years ago
Complete the equation???
soldi70 [24.7K]

Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always

Here mass number on the product side is given to us

so sum of mass number is given as

A_1 + A_2 = 265 + 1 = 266

now on the reactant side also the number must be same

A_1' + A_2' = 58 + x

now we will have

58 + x = 266

x = 208

Now number of protons on product side is given as

P_1 + P_2 = 108 + 0

now we also know that atomic number of Fe is 26

so now we will have

P_1' + P_2' = 108

26 + P_2' = 108

P_2' = 82

now the equation is given as

_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X

8 0
2 years ago
A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact
N76 [4]

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

7 0
3 years ago
Identical point charges (+50 x 10 power -6C) are placed at the corners of a square with sides of 2.0-m length. How much external
Gnesinka [82]

Answer:

636.4 J

Explanation:

The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C  and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)

Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r

U' = 4kq²/r

= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m

= 900 Nm²/√2 m

= 636.4 J

8 0
3 years ago
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