PHYSICS COLLEGE

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 1

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

Answer 2

Answer:

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

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A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest. The two blocks stick together and move with a speed of vo /3. In terms of m, mass M is most nearly_____.

Answers

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

$m_1 = m$

$m_2 = M$

$v_(1i) =v_0$

$v_(2i) = 0$

Initial momentum will be

$P_i = m_iv_(1i)+m_2v_(2i)$

$P_i = mv_0$

After collision

$v_(1f) = v_(2f) = (v_0)/(3)$

Final momentum

$P_f = (m_1+m_2)((v_0)/(3))$

$P_f = (m+M)((v_0)/(3))$

From conservation of momentum

$P_f = P_i$

Replacing,

$(m+M)((v_0)/(3))=mv_0$

$(m+M)(1)/(3) = m$

$m+M=3m$

$M=3m-m$

$M=2m$

Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Answers

Answer:

The period of the pendulum is $T = 1.68 \ sec$

Explanation:

The diagram illustrating this setup is shown on the first uploaded image

From the question we are told that

The length of the rod is $L = 80 \ cm$

The diameter of the ring is $d = 10 \ cm$

The distance of the hole from the one end $D = 15cm$

From the diagram we see that point A is the center of the brass ring

So the length from the axis of rotation is mathematically evaluated as

$AP = 80 + 10 -5 -15$

$AP = 70 \ cm = (70)/(100) = 0.7 \ m$

Now the period of the pendulum is mathematically represented as

$T = 2 \pi \sqrt{(AP)/(g) }$

$T = 2 \pi \sqrt{(0.7)/(9.8 ) }$

$T = 1.68 \ sec$

The mass of a baseball is 0.145 kg and its acceleration as it falls to the ground is 9.81 m/s2. How much force is acting on the baseball

Answers

Answer:

The answer is 1.42 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

Force = 0.145 × 9.81 = 1.42245

We have the final answer as

1.42 N

Hope this helps you

Sheba is texting and driving. She does not see the baby chinchilasitting in the road! If she was traveling at 35 m/s, and at best her
car can accelerate at -3.5 m/p/s, how far would it take her to stop?

Answers

Answer:

dfmwekognjioq 4eijqgi24gno3p2qtijoq2 gjo23ijgto32tneqew gqewkgnoweqgmnlqwe gwjoegkqwji4toqn34goti34

Explanation:

A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Answers

(a) The capacity will be "4006.4 veh/h".

(b) The speed at capacity be "25 mph".

(c) The density will be "299 veh/mi".

Given:

$q = 50k - 0.156 k^2$

At max. flow density,

$(dd)/(dk) =0$

$((dq)/(dt) ) = 50-0.321k =0$

(a)

→ $k = ((50)/(0.312) )$

$= 160.3 \ or \ 160 \ veh/mi$

By substituting the value,

→ $q = 50k-0.156k^2$

$= 50* 160.3-0.156* (160.3)^2$

$= 4006.4 \ veh/h$

(b)

The speed will be:

→ $U = (q)/(k)$

$= (4006.4)/(160.3)$

$= 25 \ mph$

(c)

The density be:

→ $1001.6 = 50k-0.156k^2$

$0.156k^2-50k+1001.6 =0$

$k = 21.5 \ veh/mi \ or \ 299 \ veh/mi$

Thus the responses above are correct.

Find out more information about density here:

brainly.com/question/6838128

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

Why must the Ojibwe people pay close attention to the seasons? a.) they must be ready to move to a new place where they can hunt

b.) they only fish during the warmest times of the day

c. they must know the right time of year for Gathering certain foods

d.) they still catch walleye with the steering method