When the sun and moon pull at right angles to the earth wat kinda tide can yu expect

Answers

Answer 1

Answer:

Answer:

A neap tide. Hope this helps

Explanation:

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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:

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.

Please provide an explanation.

Thank you!!

Answers

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

a. Left support = Right support = 22 kN

b. Left support = 36 kN

Right support = 29 kN

c. Left support force will decrease

Right support force will increase.

d. Right support = 43 kN

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)

0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support, Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)

0.6

Right support = 43 kN

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 76.2-kg water-skier has an initial speed of 5.0 m/s. Later, the speed increases to 10.4 m/s. Determine the work done by the net external force acting on the skier.

Answers

Answer:

Work done will be equal to 3186.396 J

Explanation:

We have mass m = 76.2 kg

Initial velocity u = 5 m/sec

Final velocity v = 10.4 m/sec

We have to find the work done

From work energy theorem work done is equal to change in kinetic energy

$w=(1)/(2)mv^2-(1)/(2)mu^2$

$w=(1)/(2)* 76.2* 10.4^2-(1)/(2)* 76.2* 5^2$

w = 3168.396 J

So work done will be equal to 3186.396 J

A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?

Answers

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

$v^2-u^2=2as$

$18^2-0=2* a* 125$

$a=1.296 m/s^2$

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel a distance of 125+75=200 m

$v^2-u^2=2as$

$v^2=2* 1.296* 200$

$v=√(518.4)=22.76 m/s$

You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle

Answers

Answer:

29.5 m/s

Explanation:

Volumetric flowrate = (average velocity of flow) × (cross sectional area)

Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

Cross sectional Area of flow = πr²

Diameter = 0.00579 m,

Radius, r = d/2 = 0.002895 m

A = π(0.002895)² = 0.0000037629 m²

Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)

v = 0.000111/0.0000037629

v = 29.5 m/s

Given Information:

diameter of the nozzle = d = 5.79 mm = 0.00579 m

flow rate = 0.111 liters/sec

Required Information:

Velocity = v = ?

Answer:

Velocity = 4.21 m/s

Explanation:

As we know flow rate is given by

Flow rate = Velocity*Area of nozzle

Where

Area of nozzle = πr²

where

r = d/2

r = 0.00579/2

r = 0.002895 m

Area of nozzle = πr²

Area of nozzle = π(0.002895)²

Area of nozzle = 2.6329x10⁻⁵ m²

Velocity = Flow rate/area of nozzle

Divide the litters/s by 1000 to convert into m³/s

0.111/1000 = 1.11x10⁻⁴ m³/s

Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

Velocity = 4.21 m/s

Therefore, the water exit the nozzle at a speed of 4.21 m/s

The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .

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