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astraxan [27]
3 years ago
5

Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t

he road, but then he returned to pavement. The road then turned into a gravel road. Which best describes the frictional force of his trip?
a. Friction increased when he went from pavement to ice and then decreased two more times.
b. Friction decreased when he went from pavement to ice and then increased two more times.
c. Friction increased when he went from pavement to ice and then decreased one additional time.
d. Friction decreased when he went from pavement to ice and then increased one additional time.
Physics
2 answers:
Tems11 [23]3 years ago
7 0

Answer:

b. Friction decreased when he went from pavement to ice and then increased two more times.

Explanation:

Frictional force depends on the normal force of the surface and a friction coefficient.

F_{f} = -\mu N

Since we're talking about the same car, the value of N will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.

After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

satela [25.4K]3 years ago
3 0

Answer:

b

Explanation:

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saveliy_v [14]

Answer:

The distance is 55.636 billion miles, or 528.2 AU.

Explanation:

Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

d=(20)(2781800000\ miles)=55636000000\ miles

or 55.636 billion miles.

Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

d=(55636000000\ miles)(\frac{1AU}{93000000\ miles})=598.2\ AU

6 0
3 years ago
suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain
IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

K_f = \frac{1}{2}mv^2 is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

mgh=\frac{1}{2}mv^2

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

Learn more about kinetic energy and potential energy:

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11) A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the sa
lesya692 [45]

15mph

If the man turns and runs toward point A, he will cover

3/8 of the length of the bridge in the time that it takes

the train to reach A.

If the man runs forward toward point B, what part of the bridge

will he cover before the train reaches A? Well, he will cover

3/8 of the bridge, only heading forward toward B. This will put

him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the

time the train reaches A.

since we know that the man and the train will meet at B, this

means that in the time it takes the man to run the remaining

1/4 of the bridge, the train will cover the entire length of

the bridge.

If it takes the man the same time to cover 1/4 of the bridge

that it takes the train to cover the whole bridge, then the train

must be going four times as fast as the man. Another way of saying

this is that the man runs at 1/4 the speed of the train.  

Since the train's speed is known to be 60 mph, this means that

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7 0
3 years ago
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
3 years ago
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WITCHER [35]

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA  duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

8 0
3 years ago
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