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slamgirl [31]
3 years ago
5

Shelley is in an elevator that is traveling downward and slowing down at a rate of

Physics
1 answer:
liraira [26]3 years ago
8 0

Answer:

N = 648.55[N]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

∑F =  Forces applied [N]

m = mass = 73.2 [kg]

a = acceleration = 0.950 [m/s²]

Let's assume the direction of the upward forces as positive, just as if the movement of the box is upward the acceleration will be positive.

By performing a summation of forces on the vertical axis we obtain all the required forces and other magnitudes to be determined.

-m*g + N = -m*a\\

where:

g = gravity acceleration = 9.81 [m/s²]

N = normal force (or weight) measured by the scale = 83.4 [N]

Now replacing:

-(73.2*9.81)+N=-73.2*0.950\\-718.092+N=-69.54\\N = -69.54+718.092\\N = 648.55[N]

The acceleration has a negative sign, this means that the elevator is descending at that very moment.

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someone made a copy of an analog wave in a digital wave compare the copies with the original waves and explain which type of wav
lilavasa [31]

Answer: Analog transfer information quicker, example electromagnetic waves which are used in our modern day phones.

Explanation:  Digital signals are a more reliable form of transmitting information because an error in the amplitude or frequency value would have to be very large in order to cause a jump to a different value. Signals are composed of infinite possible values. Signals are composed of only two possible values: 0 or 1.

6 0
3 years ago
Read 2 more answers
 How does ​energy change (transforms) as the mass is dropping?
olganol [36]

Answer:

Mass has total mechanical energy, which is the sum of kinetic and potential energy. as the mass is dropping, potential energy is converted into kinetic energy so mechanical energy is preserved If there is no friction. If there is friction, some of the mechanical energy is lost as heat energy so it changes.

Explanation:

4 0
3 years ago
an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco
Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

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

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

6 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth
ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

\frac{P_{1} }{\\P_2}= \frac{density * g * h_1}{density * g * h_2}

So \frac{39}{P_2}= \frac{3}{9}

Or P₂= 39 * 3 = 117 kPa

5 0
3 years ago
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