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Xelga [282]
2 years ago
12

30 points!

Physics
1 answer:
vlada-n [284]2 years ago
6 0

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc)  items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

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Hi ) how to do this??
pav-90 [236]

Answer:

X = 12N

Explanation:

Anticlockwise moment = Clockwise moment

Taking the moment at the pivot,

Anticlockwise moment = X×2.0m

Clockwise moment = (2.0N × 4.0m) + (4.0N × 4.0m)

Hence, 2X = (8 + 16)

2X = 24

X = 12N.

I HOPE IT'S OK.

4 0
3 years ago
How much work is needed to push a 117- kg packing crate a distance of 2.75 m up an inclined plane that makes an angle of 28 o wi
Nataly_w [17]

Answer:

W_{tot,min} = 1481.372\,J

Explanation:

The minimum total work is the work needed to counteract the work associated with the weight:

W_{tot, min} = W_{F}

W_{tot,min} = m\cdot g\cdot \sin \theta \cdot \Delta s

W_{tot,min} = (117\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 28^{\textdegree}) \cdot (2.75\,m)

W_{tot,min} = 1481.372\,J

6 0
3 years ago
Read 2 more answers
A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c
Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}

hence, the speed of the spheres is 37.45 m/s

8 0
3 years ago
The jumping gait of the kangaroo is efficient because energy is stored in the stretch of stout tendons in the legs; the kangaroo
Anna [14]

Answer:

the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.

w is angular frequency

formula:2π/T

now k is spring constant

F/R-->mw²

putting values:70*(2π/0.2)²

=4.9x10⁶

so we can say that SHM is not affected by the amplitude of the bounce.

6 0
3 years ago
When evaluating data, why is it better to make a graph instead of just looking at the raw data in the table?(8th grade Science s
ziro4ka [17]

Answer:All professors planned and thought about data before graph construction. When reflecting on their graphs, professors and graduate students focused on the ... feel more inclusive in the learning process and gain better science process skills, ... We organized data in a table instead of a paragraph with numbers

Explanation:

hope that helps

3 0
3 years ago
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