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natali 33 [55]
3 years ago
15

A package was determined to have a mass of 5.7 kilograms. What's the force of gravity acting on the package on earth?

Physics
2 answers:
Arturiano [62]3 years ago
8 0
1.) The force of gravity is what we call weight, we define it as:
w=mg
w=5,7kg*9,8m/s²
w=55,86kg (b)
2.) We know that:
power=W/t
power=50J/20s
power=2,5Watts (a)
3.) The work done is equal to the potential energy, so:
Epg=mgh
Epg=63kg*9,8m/s²*7m
Epg=4321J
Now we get the power:
power=W/t
power=4321J/5s
power=864Watts
Now:
1HP=746Watts
=1,16HP (b)
4.) We know that:
F=ma
350N=m*10m/s²
m=350N/10m/s²
m=35kg (b)
5.) d.) Aceleration is tha rate of change in velocity, either positive (increasing) or negative (decreasing)
Simora [160]3 years ago
7 0
==>  A package was determined . . .

                   Weight = (mass) x (gravity)

Gravity on Earth = 9.8 m/s²

                   Weight = (5.7 kg) x (9.8 m/s²)

                               =  55.86 kg-m/s² .                 1 kg-m/s² = 1 N

==> 50 J of work was performed . . .

     Power = (work done) / (time to do the work)

                 =  (50 J)  /  (20 sec)

                 =    2.5 J/sec                    1 J/sec = 1 watt

==> A 63-kg object needs . . .

       Weight = (mass) x (gravity)
       Work = (force) x (distance)
       Power = (work done) / (time to do the work)

So  Power = (mass) x (gravity) x (distance) / (time to lift)

                   = (63 kg) x (9.8 m/s²) x (7 m)  /  (5 sec)

                   =  (63 x 9.8 x 7) / (5)     (kg-m²/s²) / (sec)

                   =          864.36 joules/sec

Convert                  (864.36 watts) x (1 HP/746 watts)  =  1.159 HP

(How to remember the conversion from watts to Horsepower:
    The sails on Columbus' ship the Santa Maria developed 2 HP at top wind.
     1492 / 2  =  746 watts per HP.  )

==> A force of 350 N  causes . . .

         Force  = (mass) x (acceleration)

         Mass  =  (force) / (acceleration)

                   =  (350 kg-m/s²) / (10 m/s²)

                   =        35 kg.

==> Which of the following . . .

         Golden Rule:  "Acceleration means any change in velocity,
                                   including change in speed or direction."

A).  No.  Constant velocity means zero acceleration.
B).  No.  When people say "decelerating" they mean "slowing down".
               That's decreasing magnitude of velocity.
C). Yes.  'Acceleration' does NOT mean 'speeding up'.   Any change
               in speed or direction, even slowing down, is acceleration.
D). Yes.  If an object's velocity is changing, then either the speed or direction of its motion is changing, and either of those is 'acceleration'.  The object doesn't need to be speeding up or slowing down, but if its direction is changing, then that's called 'acceleration.  Rounding a curve at a constant speed, or driving a circle with constant speed, are both examples of 'acceleration'.
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2 years ago
A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1
baherus [9]

Answer:

The  value of the power is   P_c  =  38.55 \  W

Explanation:

From the question we are told that

   The  power  rating P_{1000} =P_b=  52 \  W

    The frequency is  f = 1000 \  Hz

    The  frequency at which the sound intensity decreases  f_k  =  20 \  Hz

     The decrease in intensity is by \beta  =  1.3 dB

Generally the  initial intensity of the speaker  is mathematically represented as

     \beta_1 =  10 log_{10} [\frac{P_b}{P_a} ]

Generally the intensity of the speaker after it has been decreased is

       \beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]

So

\beta_1-\beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]

=>  \beta =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3

=>  \beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3

=>  \beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> 10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> log_{10} [\frac{P_b}{P_c} ] = 0.13

taking atilog of both sides

[\frac{P_b}{P_c} ] = 10^{0.13}      

=>[\frac{52}{P_c} ] = 10^{0.13}      

=>  P_c  =  \frac{52}{1.34896}

=>   P_c  =  38.55 \  W

   

3 0
3 years ago
A cart on an air track is moving at 0.5m/s when the air issuddenly turned off. THe cart comes to rest in 1m. The experimentis re
Licemer1 [7]

Answer:

D = 4 m

Explanation:

Speed of cart in air track v₁ = 0.5 m/s

Speed of cart moved when air is turned off v₂= 1 m/s

The distance travelled by the cart is d₁ = 1 m

Work done (W) = F x d

Work done is equal to the kinetic energy

F x d = 1/2mv²

velocity is directly proportional to distance

therefore,

v₁²/ v₂²  = d₁ / d₂

d₂ = d₁v₂² / v₁²

= 1 m x (1 m /s)² / (0.5 m/s)²

= 4 m

5 0
3 years ago
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