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3 years ago

What is one way that friction can be reduced?

2 answers:

8 0

Friction can be reduced by making the surface smoother, lubricating the surface, or reducing the contact between the forces.

Hope this helps!

VladimirAG [237]3 years ago

6 0

Making a smoother surface for the object to move on.

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This problem has been solved!

RideAnS [48]

Answer:

Option (a)

Explanation:

We will discard options that don't fit the situation:

Option b: Incorrect since if the driver "hits the gas" then velocity is augmenting and it's not constant.

Option c and d: Incorrect since the situation doesn't give us any information that could be related directly to the terrain or movement direction.

Option a: Correct. At stage 1 we can assume the driver was going at constant speed which means acceleration is constantly zero. At stage 2 we can assume the driver augmented speed linearly, this is, with constant positive acceleration. At stage 3 we can assume the driver slowed the speed linearly, with constant negative acceleration.

6 0

3 years ago

g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward t

cupoosta [38]

Answer:

the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

Explanation:

Given that;

speed of the first spacecraft from earth v $_a$ = 0.80c

speed of the second spacecraft from earth v $_b$ = -0.60 c

Using the formula for relative motion in relativistic mechanics

u' = ( v $_a$ - v $_b$ ) / ( 1 - (v $_b$ v $_a$ / c²) )

we substitute

u' = ( 0.80c - ( -0.60c) ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )

u' = ( 0.80c + 0.60c ) / ( 1 - ( -0.48c² / c² ) )

u' = 1.4c / ( 1 - ( -0.48 ) )

u' = 1.4c / ( 1 + 0.48 )

u' = 1.4c / 1.48

u' = 0.9459c ≈ 0.95c { two decimal places }

Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

7 0

3 years ago

The wheel of a car has a radius of 20.0 cm. It initially rotates at 120 rpm. In the next minute it makes 90.0 revolutions. (a) W

just olya [345]

Answer:

Explanation:

Given that,

Radius of the wheel, r = 20 cm = 0.2 m

Initial speed of the wheel, $\omega_i=120\ rpm=753.98\ rad/s$

Displacement, $\theta=90\ rev=565.48\ rad$

To find,

The angular acceleration and the distance covered by the car.

Solution,

Let $\alpha$ is the angular acceleration of the car. Using equation of rotational kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$565.48=753.98\times 60+\dfrac{1}{2}\alpha (60)^2$

$\alpha =-24.81\ rad/s^2$

Let t is the time taken by the car before coming to rest.

$t=\dfrac{\omega_f-\omega_i}{\alpha }$

$t=\dfrac{0-753.98}{-24.81}$

t = 30.39 seconds

Let v is the linear velocity of the car. So,

$v=r\times \omega_i$

$v=0.2\times 753.98$

v = 150.79 m/s

Let d is the distance covered by the car. It can be calculated as :

$d=v\times t$

$d=150.79\ m/s\times 30.39\ s$

d = 4582.5 meters

d = 4.58 km

5 0

3 years ago

A ray of light strikes a sheet of glass (n=1.5) at an angle of 25° with the normal. Find the angle of the refracted ray within t

IRINA_888 [86]

Answer:

(d) 16°

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 25.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (glass, n=1.5)

${n_i}$ is the refractive index of the incidence medium (air, n=1)

Hence,

$1\times {sin25.0^0}={1.5}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.2817$ = 16°

7 0

4 years ago

What is the discharge of a stream that is one meter deep, two meters wide, and has a water flow of 5 meters per second

Svet_ta [14]

Answer:

10 m^3 /s

Explanation:

1 m X 2 m X 5 m /s = 10 m^3/s

4 0

2 years ago