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strojnjashka [21]
3 years ago
5

The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of th

e object on this interval by sub-dividing the interval into the indicated number of sub-intervals. Use the left endpoint of each sub-interval to compute the height of the rectangles.
v= 4t + 5(m/s) for 3 < t < 7; n = 4
The approximate displacement of the object is______m.
Physics
1 answer:
DENIUS [597]3 years ago
4 0

Answer:

The approximate displacement of the object is  <u>23  </u>m.

Explanation:

Given that:

v = 4t + 5 (m/s)  for 3< t< 7; n= 4

The approximate displacement of the object can be calculated as follows:

The velocities at the intervals of t are :

3

4

5

6

the velocity at the intervals of t =  7 will be left out due the fact that we are calculating the left endpoint Reimann sum

n = 4 since there are 4 values for t, Then there is no need to divide the velocity values

v(3) = 4(3)+5

v(3) = 12+5

v(3) = 17

v(4)= 4(4)+5

v(4) = 16 + 5

v(4) = 21

v(5)= 4(5)+5

v(5) = 20 + 5

v(5) = 25

v(6) = 4(6)+5

v(6) = 24 + 5

v(6) = 29

Using Left end point;

= \dfrac{1}{4}(17+21+25+29)

= 23 m

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In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a
timofeeve [1]

Answer:

a) F₁ = 1.48 x 10³ N

b) P = 1.88*10⁵ Pa

c) The work  is equal in both pistons

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.

Pressure is defined as the force per unit area:

P=\frac{F}{A}  Formula (1)

P₁=P₂

\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} } Formula (2)

Data

r₁= 5 cm = 0.05 m

r₂= 15 cm = 0.15 m

F₂=  13300N

Area of the pistons (A₁,A₂)

A=π*r² : Area of the circle

A₁ = π*(0.05)²=7.85*10⁻³ m²

A₂= π*(0.15)²= 70.69*10⁻³ m²

a) Force that compressed air must exert to lift a car weighing 13300 N

We replace data in the formula (2)

\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }

F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }

F₁ =  1.48 x 10³ N

b) Air pressure produced by F₁

We replace data in the formula (1)

P=\frac{F}{A}

F₁ =  1.48 x 10³ N , A₁ = 7.85*10⁻³ m²

P=\frac{1.48*10^{3} }{7.85*10^{-3} }

P= 1.88*10⁵ Pa

c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.

3 0
3 years ago
A 50-cm-long spring is suspended from the ceiling. A 270 g mass is connected to the end and held at rest with the spring unstret
AveGali [126]

Answer:

k = 22.05 N/m

Explanation:

The potential energy of the mass is converted into potential energy of the spring.

Given:

mass m = 0.27 kg

gravitational constant g = 9.8 m/s²

distance falling/ stretching of spring h = 0.24 m

U_{gravity} = U_{spring}\\ mgh = \frac{1}{2} kh^{2}

Solving for k:

k = 2mg\frac{1}{h}

5 0
4 years ago
1. A man is running on the straight road with the uniform velocity
Rufina [12.5K]
None he is undergoing constant uniform velocity. Acceleration is the rate of change of velocity
8 0
3 years ago
Read 2 more answers
6. A student travels 4.0 m east, 8.0 m south, 4.0 m west, and finally 8.0 m north. What is her
Helen [10]

The student's path traces out a rectangle and she ends up at the same spot, so her <u>displacement</u> is 0 m.

On the other hand, she travels a <u>total distance</u> equal to the perimeter of that rectangle,

4.0 m + 8.0 m + 4.0 m + 8.0 m = 24 m

3 0
2 years ago
If E1 = 13.0 V and E2 = 5.0 V , calculate the current I2 flowing in emf source E2.
swat32

The solution would be like this for this specific problem:<span>

KCL at Junction a. </span><span>
<span>+ I1 + I2 + I3 = 0 (1) </span>

<span>KVL
<span>+ 13 V - 0.2 R I1 - 0.025 R I1 - 5 V + 0.02 R I2 = 0 (2) </span>
<span>8 + 0.02 I2 = 0.225 I1 </span>
<span>I1 = 35.6 + 0.0889 I2 (2A) </span></span></span>

 

<span>KVL (bottom loop - CCW direction) </span><span>
<span>- 0.02 R I2 + 5 V + 0.5 R I3 = 0 (3) </span>
<span>0.5 I3 = -5 + 0.02 I2 </span>
<span>I3 = -10 + 0.04 I2 (3A) </span></span>

 

<span>Replace 2A and 3A into 1. </span><span>

<span>+ I1 + I2 + I3 = 0 </span>
<span>( 35.6 + 0.0889 I2 ) + I2 + ( -10 + 0.04 I2 ) = 0 </span>
<span>1.129 I2 = -25.6 </span>
<span>I2 = -22.6A </span>

<span>Solve 2A and 3A for other currents. </span>

<span>I1 = 35.6 + 0.0889 I2 = 35.6 + 0.0889 * -22.6 = 33.5A </span>
<span>I3 = -10 + 0.04 I2 = -10 + 0.04 * -22.6 = -10.9A

So the answer is letter D.</span></span>

3 0
3 years ago
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