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Strike441 [17]
4 years ago
11

Look at Concept Simulation 5.2 to review the concepts involved in this question. Two cars are identical, except for the type of

tread design on their tires. The cars are driven at the same speed and enter the same unbanked horizontal turn. Car A cannot negotiate the turn, but car B can. Which tread design, the one on car A or the one on car B, yields a larger coefficient of static friction between the tires and the road?
Physics
1 answer:
dimulka [17.4K]4 years ago
7 0

Answer:

Tread design on car B would yield a larger coefficient of static friction between the tires and the road

Explanation:

The car model using the coefficient of static friction doesn't work well with tires. A higher coefficient of static friction would require more force to cause a loss of attraction.

The static frictional force helps to keep the unbanked horizontal turn. This means that the frictional force is the centripetal force.

The tread design of car B ensures that the centripetal force is enough to negotiate the turn. On the other hand, the tread design of car A does not provide the necessary centripetal force, hence car A is unable to negotiate the turn.

Therefore, tread design on car B would yield a larger coefficient of static friction between the tires and the road.

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3 0
3 years ago
A fireman standing on a 15 m high ladder
ludmilkaskok [199]

Answer:

31.6 m/s

Explanation:

Mass is conserved, so the mass flow at the outlet of the pump equals the mass flow at the nozzle.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = 0 and P₂ = 0:

P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂

Writing v₁ in terms of v₂:

P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂

P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²

Plugging in values:

579,160 Pa − (1000 kg/m³)(9.8 m/s²)(15 m) = ½ (1000 kg/m³) (1 − (1.99 in / 3.28 in)⁴) v₂²

v₂ = 31.6 m/s

8 0
3 years ago
In Fig., block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary
Evgesh-ka [11]

Answer:

Explanation:

Given

initial height h=2.5 m

m_2=2m_1

coefficient of static friction \mu =0.5

When collision is elastic respective velocities after collision is

v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}

v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}

where u_1 and u_2=initial velocities of object

v_1 and v_2 final velocities of object

u_1=\sqrt{2\times 9.8\times 2.5}

u_1=7 m/s

v_2=\frac{0+2m_1\times 7}{m_1+2m_1}

v_2=\frac{14}{3} m/s

using v^2-u^2=2 as

0-(4.67)^2=2\times (-0.5\times 9.8)\times s

s=2.22\ m

(b)Completely Inelastic

In Completely Inelastic objects stick with each other

m_1u_1=(m_1+m_2)v

v=\frac{u_1}{3}=\frac{7}{3} m/s

using v^2-u^2=2 as

0-(2.33)^2=2\times (-0.5\times 9.8)\times s

s=0.55\ m                          

7 0
4 years ago
One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 x 10^6 Pa at 25.°C is allowed to expand t
hichkok12 [17]

Answer:

The final temperature is T2= 5.35°C

Explanation:

Apply the Gay-lussacs's law we have

\frac{P1}{T1} = \frac{P2}{T2}

P1, initial pressure= 5.00 x 10^6 Pa

T1, initiation temperature= 25.°C

P2, final pressure= 1.07 x 10^6 Pa

T2, final temperature= ?

\frac{5.00 * 10^6}{25} =\frac{1.07 *10^6}{T2} \\

Cross multiplying and making T2 subject of formula we have

T2 =\frac{1.07 *10^6*25}{5.00 * 10^6} \\\\T2= \frac{26.75}{5} \\T2= 5.35

T2= 5.35°C

6 0
3 years ago
When you don't have enough room to stop, you may _______ to avoid what's in front of you.
xxMikexx [17]
Do you have the options? I would say swerve?

7 0
3 years ago
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