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bearhunter [10]
3 years ago
7

A jogger runs at a speed of 3 m/s. How far does he run in 120 seconds?​

Physics
1 answer:
Licemer1 [7]3 years ago
8 0

Answer: *360 mph*

Explanation:

I am pretty sure that it is 360 mph

3 times 120 = 360

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Please help! Will give brainly, 50 points!! I'm stuck with this question and I don't get it!!
IrinaVladis [17]

Answer:

3.52176 x 10^-10 N

Explanation:

Fg = 3.52176 x 10^-10 Newton

8 0
3 years ago
How is energy harnessed to do useful work?
baherus [9]

Answer:

But there are ways to harness kinetic energy to either generate useful mechanical work or electricity. This is what many have tried to do to make use of energy that would be otherwise wasted. One way to harness kinetic energy that has popped up many times in recent years has to do with roads and speed bumps

Explanation:

8 0
3 years ago
Finn chooses to enter relationships only with people who share his interests in sports and work. Whenever he finds himself havin
SpyIntel [72]

Answer:

a. collectivism

Explanation:

the practice or principle of giving a group priority over each individual in it.

A good example of this is culture. Collectivism in cultural terms refers to a culture that privileges family and community over individuals. For example, children in collectivist societies are likely to take care of elderly parents if they fall ill and will change their own plans in the event of a family emergency.

8 0
3 years ago
A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he i
Tems11 [23]

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

\omega=\dfrac{2\pi \times 40}{60}\ rad/s

ω =4.18 rad/s

The radius  r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

F=22\times 4.18^2\times 1.25\ N

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

6 0
3 years ago
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
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