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3 years ago

An Object, Start from rest w Confront Aiceleration 8m/s2 along a

Physics

1 answer:

shutvik [7]3 years ago

6 0

Answer:

A) v = 40 m / s, B) v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

v = vo + a t

v = 0 + 8 5

v = 40 m / s

B) the average velocity can be found with the relation

v_average = vf + vo / 2

v-average = 0+ 40/2

v_average = 20 m / s

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The sum of the two vectors in A, B, and C is equal to the sum of the two vectors above the line. The sum of the two vectors in D isn't.

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3 years ago

When you see a straw in a glass of water it looks bent why is this ?

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3 years ago

Read 2 more answers

You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work

hammer [34]

Answer:

1) F = 71.6 N , 2) F = 120.2 N , 3) P_m= 68.600 Pa, 4) V = 2.4210-5 m³

Explanation:

This is a problem of fluid mechanics, to find the force we must use its definition

P = F / A

F = P A

The area of the circular pipe is

A = π r² = π d²/4

The pressure is given by the expression

P = P_atm + ρ g h

1) the force on the outer side is

P = P_atm

we substitute in the expression of force

F = P_atm π d² / 4

let's calculate

F = 1,013 10⁵ π 0.03²/4

F = 7.16 10¹ N

F = 71.6 N

2) the force on the inner side

the pressure

P = P_atm + ρ g h

P = 1,015 10⁵ + 1,000 9.8 7

P = 1,701 10⁵ Pa

F = 1,701 10⁵ π 0.03² / 4

F = 1,202 10²

F = 120.2 N

3) manometric pressure

Pm = ρ g h

P m = 1000 9.8 7

P_m= 68.600 Pa

4) In this part they ask for the volume that comes out in time t= 3 h

to calculate this volume we can use the flow ratio

Q = A v

V t = A v

V = A v / t

sent is the velocity of the water that comes out, to calculate it we use the Bernoulli equation

we will use index 1 for the lake surface and ionice 2 apa the position of the plug

P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

As the lake has much more capacity than the pipeline, the velocity of the surface of the lake is peeling, in this case we approximate it steel

(P₁-P₂) + ρ G (y₁ -y₂) = ρ g v₂²

1000 9.8 v₂² = ρ g h + 1000 9.8 (7-0)

9800 v₂² = 1000 9.8 7 + 68600

v₂ = √ (137200)

v₂ = 370.4 m / s

t = 3 h (3600s / h) = 10800 s

we substitute in the volume equation

V = π d²/4 370.4 / 10800

V = 2.4210-5 m³

4 0

3 years ago

You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an

Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

7 0

3 years ago

A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at

fenix001 [56]

Answer:

a) $I_{RMS} = 4.79 A$

b) $PF = 0.908$

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, $cos \theta_{1} = 0.92$

$\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074$

$Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W$

For load 2

Active power, P₂ = 250 W

Power factor, $cos \theta_{2} = 0.8$

$\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87$

$Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W$

For load 3

Active power, P₃ = 250 W

Power factor, $cos \theta_{3} = 1$

$\theta_{3} = cos^{-1} 1\\\theta_{3} =0$

$Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W$

Calculate the total reactive power, $Q_{net} = 42.6 + 187.5 + 0$

$Q_{net} = 230.1 W$

Calculate the total active power, $P_{net} = 100 + 250 + 150 = 500 W$

$S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1$

$P_{net} = IVcos \theta_{net}$

$\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712$

V = 115 $V_{rms}$

$500 = I_{RMS} * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A$

b) Power factor of the composite load is $cos\theta_{net}$

$\theta_{net} = 24.712\\PF = cos 24.712\\PF = 0.908$

4 0

3 years ago