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makvit [3.9K]
4 years ago
14

"How much work is required to lift an object with a mass of 5.0 kilograms to a height of 3.5 meters?" What would be the formula

for this?
Physics
1 answer:
grandymaker [24]4 years ago
4 0
The answer is 172 J (or 1.72 · 10² J)

A work (W) is a product of a force (F) and a distance (d): W = F · d
Since the force is a product of mass and acceleration: F = m · a
then work would be: W = m · a · h

It is known:
m = 5 kg
h = 3.5 m
a = 9.8 m/s² (gravitational acceleration, since is expected that on some height gravitation force will work).

Therefore,
W = 5 · 9.8 · 3.5 = 171.5 J ≈ 172 J ≈ 1.72 · 10² J
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In a mass spectrometer, a single-charged particle (charge e) has a speed of 1.0 × 10 6 m/s and enters a uniform magnetic field o
Nonamiya [84]

Answer:

The mass is  m  =6.4*10^{-28} \ kg

Explanation:

From the question we are told that

   The  speed of the charge is  v   = 1.0 *10^{6} \  m/s

    The  magnetic field is  B = 0.20 \ T

     The radius is r  =  0.02 \ m

      The value of the charge is  e  = 1.60 *10^{-19} \  C

The centripetal acting on the charge moving in the circular orbit is mathematically represented as

        F_c  =  \frac{mv^2}{r }

Now this centripetal force is due to the force exerted on the charge by the magnetic field on the charge which is mathematically represented as

     F_m  =  qv  B  sin\theta

At the maximum of this magnetic force \theta =  90 ^o

So  

     F_m  =  e v  B  sin(90)

      F_m  =  e v  B

Now given that it is this  magnetic force that is causing the circular motion we have that

       F_c  =  F_m

=>     \frac{mv^2}{r }  =  ev  B

=>     m  = \frac{e * B  *  r  }{v }

substituting values

       m  = \frac{ 1.60 *10^{-19} *  0.20   *  0.020   }{1.0*10^{6} }

     m  =6.4*10^{-28} \ kg

8 0
4 years ago
A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another
Alex_Xolod [135]

Answer:

15.8 ft/s

Explanation:

\frac{da}{dt} = Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

\frac{db}{dt} = Velocity of car B = 13 ft/s

b = Distance car B travels = ft

c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

6 0
4 years ago
when a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of
matrenka [14]
1 volt. 
1/1 = 1
2/2 = 1
7 0
3 years ago
The greater the difference in electronegativity between two covalently bonded<br><br> atoms
katrin [286]

Answer:

The greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

Explanation:

Bond polarity (i.e the separation of electric charge along a bond) and ionic character (amount of electron sharing) increase with an increasing difference in electronegativity.

Therefore, we can say that, the greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

7 0
3 years ago
Sound travels at 343 m/s through dry air. If a lightening bolt strikes the ground 2 km away from you, how long will it take for
Minchanka [31]

Answer:

0.006 km / m/s

Explanation:

divide 343 m/s and 2 km

4 0
2 years ago
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