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Neporo4naja [7]
3 years ago
13

What is the speed of a walking person in m/s if the person travels 1000 m in 1200 seconds?​

Physics
1 answer:
Brilliant_brown [7]3 years ago
4 0

Answer:

0.83

Explanation:

to figure out what m/s is you would divide the distance by time.

1000 divided by 1200 is equal to .83  so

.83 m/s as your final answer

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The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

\Sigma F = F_{E}-W = 0 (Eq. 1)

Where:

F_{E} - Electrostatic force exerted on human, measured in Newton.

W - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

q\cdot E-m\cdot g = 0

q = \frac{m\cdot g}{E} (Eq. 2)

E - Electric field, measured in Newtons per Coloumb.

m - Mass, measured in kilograms.

g - Gravity acceleration, measured in meters per square second.

q - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that m = 60\,kg, g = 9.807\,\frac{m}{s^{2}} and E = -150\,\frac{N}{C}, the charge that a 60-kg human must have to overcome weight is:

q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }

q = -3.923\,C

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

F = \kappa \cdot \frac{q^{2}}{r^{2}} (Eq. 3)

Where:

\kappa - Electrostatic constant, measured in Newton-square meter per square Coulomb.

q - Electric charge, measured in Coulomb.

r - Distance between two people, measured in meters.

If we know that \kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q = -3.923\,C and r = 100\,m, then the force of repulsion between two people is:

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]

F = 13.851\times 10^{6}\,N

The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0
3 years ago
Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0
3 years ago
Considering the factors that affect gravitational pull, in which location would the gravitational pull from the earth be SMALLES
OleMash [197]
C. The higher the altitude the less gravity affects you
4 0
3 years ago
Spidermans nemesis electro delivers 4kj of electrical energy in half a second how much power does it draw from the mains?
Elodia [21]

Power delivered = (energy delivered) / (time to deliver the energy)

Power delivered = (4,000 J) / (0.5 sec)

Power delivered = 8,000 watts

I'm a little surprised to learn that Electro draws his power from the mains.  This is VERY good news for Spiderman !  It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.

But OK.  Let's assume that Electro draws it all from the mains.  Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).

The efficiency of Electro's internal process is

<em>(power he shoots out) / (power he draws from the mains) </em>.

So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of

<em>(8,000 watts) / (his internal efficiency) .  </em>

4 0
3 years ago
Read 2 more answers
Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at
Karo-lina-s [1.5K]
Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.

Speed will be = squreroot ( 2*g*L)

L is length of pendulum
5 0
4 years ago
Read 2 more answers
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