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cupoosta [38]
2 years ago
7

The kinetic energy of a car is 8 × 106 J as it travels along a horizontal road. How much power is required to stop the car in 10

s?
Physics
1 answer:
Ratling [72]2 years ago
7 0

Answer:

<em>The power required to stop the car =  8 × 10⁵ watt.</em>

Explanation:

Kinetic Energy: This is the energy of a body in motion. The S.I unit of energy is Joules (J).

Power: These is defined as time rate of doing work, or it can be defined as the time rate of transfer of energy. The S.I unit of power is Watt (W).

Mathematically, power can be expressed as

Power = Energy/time

P = E/t................. Equation 1

<em>Given: E = 8×10⁶ J, t = 10 s</em>

<em>Substituting these values into equation 1,</em>

<em>p = 8×10⁶/10</em>

<em>p = 8 × 10⁵ watt.</em>

<em>Therefore the power required to stop the car =  8 × 10⁵ watt.</em>

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2 years ago
What is the direction of the normal contact force of the road on the wheels?
Gekata [30.6K]

Answer:

The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.

Explanation:

3 0
2 years ago
g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t
jok3333 [9.3K]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}            (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

v_{max}=\omega A=2\pi f A      (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

a_{max}=\omega^2A=(2\pi f)^2 A

a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J

The total energy is 0.1J

(e) The displacement as a function of time is:

x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)

6 0
2 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
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