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mamaluj [8]
3 years ago
11

Where are you atera11

Physics
1 answer:
Andre45 [30]3 years ago
8 0

Answer:

? who is atera11?

Explanation:

You might be interested in
Two identical charged pith balls are brought together to touch each other. They are then
sergejj [24]

Answer:

-17.5 nC

Explanation:

charge A = -30 nC

charge B = -5 nC

After adding them it would be the average of the two charges because of the getting same voltage difference. so

c = (-30+(-5)) / 2 nC

c= -17.5 nC

answer is -17.5 nC

4 0
3 years ago
26. An ice-skater who weighs 200 N is gliding across the ice. If the force of friction is 4 N. what is the
Scrat [10]

Answer:

0.02

Explanation:

coefficient of kinetic friction = μ

force of friction = Ff

Normal Force = FN, but

FN = -W

Ff = -μFN

so μ = Ff/FN

= 4N/200N

= 0.02.

7 0
3 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender
aalyn [17]

1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.

2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.

 

<u>Justification:</u>

There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors.  The reliable and accurate network environment makes a single frame economically better.

Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.

With encapsulation, each layer:

  • provides a service to the layer above it
  • communicates with a corresponding receiving node

Thus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead.  This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.

Learn more about encapsulation of packets here: brainly.com/question/22471914

6 0
2 years ago
A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.
horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

7 0
2 years ago
Read 2 more answers
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