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EastWind [94]
3 years ago
8

A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca

r ?
Physics
1 answer:
natka813 [3]3 years ago
7 0

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
3 years ago
A diagram of a closed circuit with a power source on the left labeled 6 V. There are 3 resistors in parallel, separate paths, co
AfilCa [17]

Answer:

Current: 1.0 Amperes

The minimum current is flowing through path D

Explanation:

We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

\frac{1}{R_e} =\frac{1}{R_B}+\frac{1}{R_C}+\frac{1}{R_D}\\\frac{1}{R_e} =\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17\\R_e=(1/0.17)\Omega\\R_e=5.88 \Omega

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

V=I*R\\6V=I*5.88\Omega\\I=\frac{6}{5.88} Amp\\I=1.02A

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.

The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.

Than means that the lowest current will be registered through branch D where the 50 \Omega resistor is.

8 0
2 years ago
Read 2 more answers
In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s
serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

a = \dfrac{1.2 \times 10^5}{86}

a =1395.35\ m/s^2

Using equation of motion

v² = u² + 2 a s

s =\dfrac{v^2}{2a}

s =\dfrac{52^2}{2\times 1395.35}

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

8 0
3 years ago
What would occur if the index of refraction of a material was less than n=1?
Bingel [31]

The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.

The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.

n=C/V

here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.

Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.


3 0
3 years ago
Read 2 more answers
A 2.00 kg ball is thrown upward at Some unknown angle from the top of a 20.0 M high building If the initial magnitude of the vel
mixer [17]

Answer:

792 J

Explanation:

The total energy of the ball is E = U + K where U = potential energy = mgh and K = kinetic energy = 1/2mv²

E = mgh + 1/2mv² where m = mass of ball = 2.0 kg, g = acceleration due to gravity = 9.8 m/s², h = height of building = 20.0 m, v = initial velocity of ball = 20.0 m/s.

So, substituting the values of the variables into E, we have

E = mgh + 1/2mv²

= 2.00 kg × 9.8 m/s² × 20.0 m + 1/2 × 2.00 kg × (20.0 m/s)²

=  392 J + 400 J

= 792 J

6 0
2 years ago
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