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MAVERICK [17]
3 years ago
9

Consider your desk at rest on your bedroom floor. As you and your friend start to lift it, does the support force on the desk pr

ovided by the floor increase, decrease, or remain unchanged? What happens to the support force on the feet of you and your friend?
Physics
1 answer:
Virty [35]3 years ago
7 0

A) From rest and before performing the described exercise, the entire weight of the desk is supported by the floor. This support force (normal force) exerted by the floor will be equivalent to the force equivalent to the weight of the desk. From the moment the desk starts to be lifted, said support force will decrease and the weight of the desk supported by the floor will be less.

B) As the desk is lifted, the support force of the two guests will increase, this occurs because the force is transferred to the support of both, but the support force of the floor on the two people will increase.

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while flying a plane parallel to the ground a pilot releases a fuel tank in order to reduce the planes mass. what is the tanks f
stepan [7]

La velocidad vertical del tanque después de caer 10 m es 14 m/seg .

La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :

Vfy=?

  h = 10 m

                              Fórmula de Velocidad vertical Vfy:

                           Vfy²  = 2*g*h

                            Vfy= √(2*9.8m/seg2* 10m )

                            Vfy= 14 m/seg

7 0
3 years ago
A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

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

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

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

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
This list shows the heights in feet of the 30 tallest mountains in the United States. 14,269 14,361 14,470 14,831 16,237 14,270
antoniya [11.8K]

Answer: a. Frequency. 20, 4, 3, 1, 1, 0, 1

Explanation:

Interval(Height in feet) - - - - - - - - frequency

14000 - 14999 - - - - - - - - - - - - - - - 20

15000 - 15999 - - - - - - - - - - - - - - - 4

16000 - 16999 - - - - - - - - - - - - - - - 3

17000 - 17999 - - - - - - - - - - - - - - - 1

18000 - 18999 - - - - - - - - - - - - - - - 1

19000 - 19999 - - - - - - - - - - - - - - - 0

20000 - 20999 - - - - - - - - - - - - - - 1

Frequency : (20, 4, 3 1, 1, 0, 1)

8 0
3 years ago
g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin
Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

       i.e Normal \ Force(N)  = mg cos \theta

And \theta = 90 so

           N = 0

c

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position        

d

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work  

Explanation:

5 0
3 years ago
Read 2 more answers
You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear
Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

r_2 =15 \times 10^{-1.5}

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0
3 years ago
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