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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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HELP ASAP PLEASE!!!!!!!
pshichka [43]

Answer:a

Explanation: the plates shifts cracking making earthquakes

8 0
3 years ago
__________ is when a person becomes less sensitive to a drug. A. Tolerance B. Craving C. Addiction D. Dependence
Luden [163]
A) Tolerance

Tolerance is developed after using a drug repeatedly, so the body adapts to it. Because of that, people who develop a tolerance would then need to use more of that drug to get the same effect.
4 0
3 years ago
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , suc
marysya [2.9K]

Answer:

    q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

                    ΔV = ΔV_{C}

we assume that the source has a voltage of the form

                    ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

                   C = q / ΔV

                  q = C ΔV sin wt

the current in the circuit is

                    i = dq / dt

                    i = c ΔV₀ w cos wt

if we use

                  cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

                  i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

                     V_{L} + V_{C} = 0

                      L di / dt + q / C = 0

the current is

                      i = dq / dt

                       

they ask us for a solution so that

                    L d²q / dt² + 1 / C q = 0

                     d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

                    q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

                di / dt = Aw cos (wt + Ф)

                d²i / dt²= - A w² sin (wt + Ф)

                 - A w² + 1 /LC  A = 0

                  w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

                 0 = A sin Ф

                  Ф = 0

             

                  q = q₀ sin (wt)

6 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
mixer [17]

The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

7 0
3 years ago
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