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You're driving along at 25 m/s with your aunt's valuable antiques in the back of your pickup truck when suddenly you see a giant hole in the road 55 m ahead of you. Fortunately, your foot is right beside the brake and your reaction time is zero! Can you stop without the antiques sliding and being damaged? Their coefficients of friction are μs=0.6 and μk=0.3. Hint: You're not trying to stop in the shortest possible distance. What's your best strategy for avoiding damage to the antiques?

Answers

Answer 1
Answer:

Answer:

Explanation:

initial velocity, u = 25 m/s

distance, s = 55 m

coefficient of static friction = 0.6

coefficient of kinetic friction = 0.3

Let the acceleration is a.

Use third equation of motion

v² = u² + 2as

0 = 25 x 25 - 2 x a x 55

a = 5.68

a = μg

μ = 5.68 / 9.8 = 0.58

so, the coefficient of friction is less then the coefficient of static friction so the antiques are safe.

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How does the geosphere interact with the hydrosphere

Answers

 Plants (biosphere) draw water (hydrosphere) and nutrients from the soil (geosphere) and release water vapor into the atmosphere.

The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.

Answers

Answer:

d = 2*0.87 = 1.75 cm

Explanation:

by using flow rate equation to determine the  speed in larger pipe

\phi =\pi r^2 v

v = (\phi)/(\pi r^2)

  = (2900 cm^3/s)/(3.14(1.25cm)^2)

= 591.10 cm/s

 = 5.91 m/s

by Bernoulli's EQUATION

p1 +(1)/(2) \rho v1^2 = p2 +(1)/(2) \rho v2^2

139000+ (1)/(2)*1000*5.91^2 = 101000 +(1)/(2)*1000* v2^2

solving for v2

v2 = 10.53 m/s

diameter can be determine by using flow rate equation

q = v \pi r^2

r^2 = (q)/(\pi v)

     = (2900)/(3.14*1053)

r = 0.87 cm

d = 2*0.87 = 1.75 cm

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

Answers

Answer

given,                                                

Tension of string is F                                                  

velocity is increased and the radius is not changed.      

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone          

  = (mv^2)/(r)                          

now centrifugal force is balanced by tension

T =(mv^2)/(r)                                

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upwards a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block

Answers

Answer:

423m/s

Explanation:

Suppose after the impact, the bullet-block system swings upward a vertical distance of 0.4 m. That's means their kinetic energy is converted to potential energy:

E_p = E_k

mgh = mv^2/2

where m is the total mass and h is the vertical distance traveled, v is the velocity right after the impact at, which we can solve by divide both sides my m

Let g = 9.81 m/s2

gh = v^2/2

v^2 = 2gh = 2 * 9.81* 0.4 = 7.848

v = √(7.848) = 2.8m/s

According the law of momentum conservation, momentum before and after the impact must be the same

m_uv_u + m_ov_o = (m_u + m_o)v

where m_u = 0.01, v_u are the mass and velocity of the bullet before the impact, respectively.m_ov_o are the mass and velocity of the block before the impact, respectively, which is 0 because the block was stationary before the impact

0.01v_u + 0 = (0.01 + 1.5)*2.8

0.01v_u = 4.23

v_u = 4.23 / 0.01 = 423 m/s

A spectroscope:measures light from distant objects
makes object far away look closer
receives radio signals from objects in space

Answers

Answer:

Option A

Measures light from distant objects

Explanation:

A spectroscope is used to measure the use of light from a distant object to work out the object is made of.

It could be the single-most powerful tool astronomers use.

Professor Fred Watson from the Australian Astronomical Observatory says that "It lets you see the chemicals being absorbed or emitted by the light source"

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = \sqrt{3^(2) +2^(2) } = √(13) = 3.61cm = 0.036m

r₂ = \sqrt{4^(2) + 3^(2) } = √(25) = 5cm = 0.05m

electric potential V = (kq)/(r)

change in potential ΔV = V_(1) - V_(2)

ΔV = (2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) ) , where q_(1) = q_(2)=2.00μC

ΔV = 2kq((1)/(r_(1)) - (1)/(r_(2) ))

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × ((1)/(0.036) - (1)/(0.05) )

ΔV= 2.789×10⁵

(1)/(2)mv^(2) = ΔV × q₃

(1)/(2) ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561
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