PHYSICS COLLEGE

You are exiting a highway and need to slow down on the off-ramp in order to make the curve. It is rainy and the coefficient of static friction between your tires and the road is only 0.4. If the radius of the off-ramp curve is 36 m, then to what speed do you need to slow down the car in order to make the curve without sliding?

Answers

Answer 1
Answer:

Answer:

11.87m/s

Explanation:

To solve this problem it is necessary to apply the concepts related to frictional force and centripetal force.

The frictional force of an object is given by the equation

F_r = \mu N

Where,

\mu =Friction Coefficient

N = Normal Force, given also as mass for acceleration gravity

In the other hand we have that centripetal force is given by,

F_c= (mv^2)/(R)

The force experienced to stay on the road through friction is equal to that of the centripetal force, therefore

F_r = F_c

\mu mg = (mv^2)/(R)

Re-arrange to find the velocity,

V = √(R\mu g)

V = √(36*0.4*9.8)

V = 11.87m/s

Therefore the speed that it is necessaty to slow down the car in order to make the curve without sliding is 11.87m/s

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A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?

Answers

I can't give you the actual number of turns, because it's the RATIO
that counts. 

However many turns the primary has, the secondary should have
about TEN TIMES that number.  Then the transformer will multiply
the primary voltage by 10 ... 120 volts of AC at the primary will
become 1,200 volts of AC at the secondary.

Note that it HAS TObe AC.  If the transformer is supplied with DC,
then 120 volts at the primary becomes zero volts at the secondary
and a big cloud of stinky smoke in the room.

What is the magnitude of the force you must exert on the rope in order to accelerate upward at 1.4 m/s2 , assuming your inertia is 63 kg ? Express your answer with the appropriate units.

Answers

Answer:

The magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Explanation:

The magnitude of force, you must exert can be estimated as follows;

Since it is upward motion, we must consider acceleration due to gravity which opposes the upward motion.

F = m(a+g)

where;

F is the magnitude of the upward force

m is your mass, which is the measure of inertia = 63kg

a is the acceleration of the rope = 1.4 m/s²

F = 63(1.4 + 9.8)

F = 63(11.2)

F = 705.6 N

Therefore, the magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Answer:

705.6 N

Explanation:

Force: This can be defined as the product of mass a acceleration.

The S.I unit of force is Newton.

The expression for the force on the rope in order to accelerate upward is given as,

F-W = ma .......................... Equation 1

Where F = Force exerted on the rope, W = weight of the rope, m = mass of the rope, a = acceleration.

But,

W = mg........................ Equation 2

Where g = acceleration due to gravity

substitute equation 2 into equation 1

F-mg = ma

F = ma+mg

F = m(a+g).............. Equation 3

Given:  m = 63 kg, a = 1.4 m/s²

Constant: g = 9.8 m/s²

Substitute into equation 3

F = 63(1.4+9.8)

F = 63(11.2)

F = 705.6 N

The magnitude of the force exerted on the rope = 705.6 N

Two microwave frequencies are authorized for use in microwave ovens: 895 and 2560 mhz. calculate the wavelength of each.

Answers

Wavelength = (speed) / (frequency)

The speeds of the two possible signals are equal, just like
all other forms of electromagnetic radiation.

Wavelength of 895 MHz = (3 x 10⁸ m/s) / (8.95 x 10⁸/s) = 0.335 m

Wavelength of 2560 MHz = (3 x 10⁸ m/s) / (2.56 x 10⁹/s) =  0.117 m 

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

The formula for the rotational kinetic energy is

KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^(2)=(2.67 \ kg) (0.777 \ m)^(2) =2.07459 \ kgm^(2)

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_(rot) =( (1)/(2) )(5)(2.07459 \ kgm^(2)) (60\ rad/s)^(2)

KE_(rot) =18,671.31 \ J

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> 573*2\pi *(1)/(60) =60rad/s

I=(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537kgm^2

K=1/2(number of rods)(I)(ω)=(1)/(2) *(5)(0.537)(60)^2=4833J

I know it's very late, but hope this helps anyone else trying to find the answer.

On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0ºC and completely melts to 0ºC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s) ?

Answers

Answer:

P=13.5 W

Explanation:

In this case, power is the rate of transferring heat per unit time:

P=(Q)/(\Delta t)(1)

The heat is given by the formula of the latent heat of fusion, since the ice is melting.

Q=mL_f(2)

Here m is the ice's mass and L_f is the heat of fusion of ice. Recall that one day has 86400 seconds. Replacing (2) in (1) and solving:

P=(mL_f)/(\Delta t)\nP=(3.5kg(334*10^3(J)/(kg)))/(86400s)\nP=13.5 W
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