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Anastaziya [24]
3 years ago
14

Help me to answer my questions please

Physics
1 answer:
Andrei [34K]3 years ago
3 0
2 is c
3 is a 
4 is b
5 is c
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What do you think will happen to the size of the splat of an egg as we increase in height ?
yawa3891 [41]

Answer:

It will be more spread out

Explanation:

As you increase the height, the gravitational potential energy increases and it gets faster, causing it to hit the ground harder. This in turn makes the 'splat' be more spread out

4 0
3 years ago
Please help...........................
Marianna [84]
Hi!

Neutrons are neutral, which means they don't exactly have an electrical charge. It's because of this neutral charge that it is represented with a '0'. 

On the other hand, protons and electrons <em>do </em>have electrical charges. Electrons flow around the outside of the nucleus, with a negative charge.

Protons are stored in the nucleus with the neutrons, holding a positive charge.

Hopefully, this helps! =)
6 0
4 years ago
Read 2 more answers
A conductor shaped as a circular loop with a radius of 4.0 m is located in a uniform but changing magnetic field. If the maximum
bekas [8.4K]

Answer:

\frac{\delta B}{\delta t}= 0.0995 \  T/s

Explanation:

Given that :

The radius of the circular loop = 4.0 m

Maximum Emf E_{max} = 5.0 V

The  maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies can be determined via the expression;

E_{max} = Area (A) * \frac{\delta B}{\delta t}

E_{max} = \pi r^2 * \frac{\delta B}{\delta t}

5.0 = \pi * (4.0)^2 * \frac{\delta B}{\delta t}

5.0 = 50.27 * \frac{\delta B}{\delta t}

\frac{\delta B}{\delta t}= \frac{5}{50.27}

\frac{\delta B}{\delta t}= 0.0995 \  T/s

5 0
3 years ago
The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10
Ilya [14]

The solution is in the attachment

4 0
3 years ago
Read 2 more answers
A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the
OLga [1]

Answer: a) -127 N b) No. c) -112 N  d) 40 N e) 15 N

Explanation:

a) Friction force always oppose to the relative movement between two surfaces, and, provided that be less than the fs max, adopt any value to counteract the applied force.

The fs max, is the horizontal component of the contact force, and can be written as follows:

Fs max = us . Fn  

As the block is at rest in the vertical direction, this means that Fn must be numerically equal to the weight of the object:

Fn = m g = 35 kg. 9.8 m/s2 = 343 N → Fs max = 0.37. 343 N = 127 N

b) Now, as the applied force is smaller than Fs max, this means that the friction force, is equal and opposite to the applied forcé, i.e., -112 N, so the crate doesn´t move.

c) Please see above.

d) As explained above, the maximum friction force, is proportional to the normal force, which adopts any value needed to satisfy the Newton´s 2nd Law.

So, if we diminish the normal force, we can lower the máximum friction force, helping to the worker to move the crate.

The mínimum needed normal force, will be the one that satisfies the following:

Fs max = F applied = us Fn = 112 N = 0.37. Fn.

Solving for Fn, we get Fn= 303 N

So, the difference between the original normal force and the new one, will be the mínimum upward force needed to make the crate to move, as follows:

Fup = 343 N -303 N = 40 N  

e) If we keep the normal force unchanged, but add an horizontal force to help the worker, we will need that the sum of both forces, will be equal to the Fs max, as follows:

Fh + Fapp = 127 N → Fh = 127 N – 112 N = 15 N

5 0
3 years ago
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