1. A bicyclist starts at 2.5 m/s and accelerates along a straight path to a speed of 12.5 m/s ina time of 4.5 seconds. What is the bicyclist's acceleration to the nearest tenth of a m/s??

Answers

Answer 1
Answer:

Answer:

2.2m/s

Explanation:

a=v-u/t

12.5-2.5/4.5=2.222

~2.2m/s


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An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force .
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When you blow some air above a paper strip, the paper, rises. This is because 1.) the air above the paper moves faster and the pressure is lower 2.) the air above moves faster and the pressure is higher 3.) the air above the paper moves faster and the pressure remains constant 4.) the air above the paper moves slower and the pressure is higher 5.) the air above the paper moves slower and the pressure is lower

A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

Answers

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies  in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3

D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3

D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0)=88  BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1            +      m2       =  m3

-

0.376m1   +  1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for  2" Sch 40=2.067in=0.17225ft

A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6.use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Answers

The baseball curves better at a flatplain due to contacting with air.

What is a baseball curve?

A curveball is a breaking pitch with more movement than most other pitches. It is thrown slower and with more overall break than a slider and is used to throw hitters off balance.

On a flat plain, a baseball will curve down better. This is due to the curving being caused by the ball contacting air and being pushed in a specificdirection.

The spin, speed, and location of the ball's stitches in relation to the air will all influence how it changes direction when pushed against.

Consider throwing a baseball in a vacuum where there is no air. Because there is no air to push on the ball, it will not curve at all.

Thus, a flat plain area will be better for baseball curve.

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A baseball will curve better down on a flat plain.In view of the fact that the curving is caused by the ball contacting the air and pushing the ball in a particular direction.

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

Answers

B. 60 cm 

  All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

A cylinder with a diameter of 2.0 in. and height of 3 in. solidifies in 3 minutes in a sand casting operation. What is the solidification time if the cylinder height is doubled? What is the time if the diameter is doubled?

Answers

Answer:

3 min 55 sec is the solidification time if the cylinder height is doubled

7min 40 sec if the diameter is doubled

Explanation:

see the attachment

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Answers

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy,  U = 13.0 J

The volume,  V = 6.00 mm^3 = 6.00*10^(-9)m^3

The potential energy per unit volume is defined as the energy density.

u = (U)/(V)

u= ((13.0 J))/((6.00*10^(-9) m^3))

u= 2.167109 J/m^3

The energy density related with electric field is given by

u = (1)/(2) \epsilon_0 E^2

Here, the permitivity of the free space is

\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2

Therefore, rerranging to find the electric field strength we have,

E = \sqrt{(2u)/(\epsilon_0)}

E = \sqrt{(2(2.167109))/(8.85*10^(-12))}

E = 2.211010 V/m

Therefore the electric field is 2.21V/m

Final answer:

To calculate the electric field strength that would store 12.5 Joules of energy in every 6.00 mm^3 of space, we use the energy density formula. We firstly find the energy density and input it into the formula to solve for the electric field strength. The result is approximately 6.87 X 10^6 N/C.

Explanation:

The energy stored in an electric field is given by the formula U = 1/2 ε E^2. Here, U is the  energy density (energy per unit volume), E is the electric field strength, and ε is the permittivity of free space.  

Given that the energy stored U is 12.5 joules, and the volume is 6.00 mm^3 or 6.00X10^-9 m^3, the energy density (U) can be computed as 12.5 J/6.00X10^-9 m^3 = 2.08X10^12 Joule/meter^3.

We can solve the formula for E (electric field strength): E = sqrt ((2U)/ε). Substituting the value of ε (8.85 × 10^-12 m^-3 kg^-1 s^4 A^2), we can find E to be approximately 6.87 X 10^6 N/C.

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You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

Answers

Answer:

d'=√(v^2+w^2)

Explanation:

Rate of Change

When an object moves at constant speed v, the distance traveled at time t is

x=v.t

We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

y=v.t

and the horizontal distance is

x=w.t

The distance between both friends is computed as the hypotenuse of the triangle

d^2=x^2+y^2

We need to find d', the rate of change of the distance between both friends.

Plugging in the above relations

d^2=(v.t)^2+(w.t)^2

d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2

Solving for d

d=√((v^2+w^2)t^2)

d=√((v^2+w^2)).t

Differentiating with respect to t

\boxed{d'=√(v^2+w^2)}

Final answer:

The problem is solved using Pythagoras' Theorem, representing the two travel paths forming a right triangle. The rate at which the distance increases between two points moving perpendicularly can be found by differentiating the resulting equation, which yields the expression sqrt[(v^2)+(u^2)].

Explanation:

The question is about the rate at which the distance between you and your friend is increasing at time t. It's a typical problem in kinematics. Because the roads are perpendicular to each other, we can solve the problem using Pythagoras' Theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's denote the distance you've traveled as D1 = v*t (because distance = speed * time) and the distance your friend has travelled D2 = u*t. The distance between you can be computed using Pythagoras' Theorem as D = sqrt(D1^2 + D2^2). Hence, D = sqrt[(v*t)^2 + (u*t)^2]. Differentiating D with respect to t using the chain rule will give us the rate at which the distance between you is increasing, which is sqrt[(v^2)+(u^2)].

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