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ICE Princess25 [194]
3 years ago
15

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

m necessary. find the magnitude of the frictional force between the person and the wall.
Physics
2 answers:
Naddika [18.5K]3 years ago
7 0
In the magnitud is in good
Kamila [148]3 years ago
7 0

Answer:

friction force must be equal to the magnitude of weight of the object

Explanation:

When the object is stuck against the wall

then we will have

F_n = \frac{mv^2}{r}

also we know that

mg = F_f

so if the block is at rest and not sliding then in vertical direction the force must be balanced

so here we can say

F_g = F_f

so friction force must be equal to the magnitude of weight of the object

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0
3 years ago
Please help me someone !
san4es73 [151]

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

7 0
2 years ago
Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s
Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19=  -5.625 10^35 eV

the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
8 0
3 years ago
True or false money difficult to carry around​
Nina [5.8K]

Answer

yeard

Explanation:

cause it is

6 0
3 years ago
Read 2 more answers
How is the speed of an object related to the amount of<br> kinetic energy the object has?
xenn [34]

Answer:

directly proportional to the square of its speed.

Explanation:

;)

3 0
2 years ago
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