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3 years ago

How are acceleration and velocity related

Physics

1 answer:

IRINA_888 [86]3 years ago

5 0

Answer:

answer is C

Explanation:

acceleration is rate of changing velocity in a time

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A 1.1-cm-diameter pipe widens to 2.34 cm, then narrows to 4.9 mm. Liquid flows through the first segment at a speed of 3.57 m/s.

ratelena [41]

Answer:

v2=0.79 m/s

Explanation:

equation of continuity: V1A1=V2A2

A=(pi)r^2

A1=(pi)(0.55cm)^2

A2=(pi)(1.17cm)^2

convert cm2->m2

A1= 0.95cm2=

A2= 4.3cm2=

(3.57 m/s)(0.000095 m^2)=(x m/s)(0.00043 m^2)

v2=0.79 m/s

hope this helped

6 0

3 years ago

For what value of the ratio r/a of plate radius to separation between the plates does the electric field at the point x=a/2 on t

SIZIF [17.4K]

solution:

E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})

x=\frac{r}{a}

infinite case,

Ei=\frac{r}{\epsilon0}

\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})

we have to find x when,

ei-e\delta =1% ,y=ei=1/100 ei

or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei

\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}

4x^2+1 =10^4

x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50

\therefore \frac{r}{a}\approx 50

6 0

3 years ago

With what tension must a rope with length 2.90 m and mass 0.125 kg be stretched for transverse waves of frequency 42.0 Hz to hav

Vesnalui [34]

Answer:

41.64 N

Explanation:

Applying,

v = √(T/m')................ Equation 1

Where v = velocity of the wave, T = Tension of the rope, m' = mass per unit length of the rope.

make T the subject of the equation,

T = v²m'................. Equation 2

But,

v = λf............... Equation 3

Where λ = wavelength, f = frequency

And

m' = m/L........... Equation 4

Where m = mass of the rope, L = length of the rope

Substitute equation 3 and equation 4 into equation 2

T = (λf)²(m/L).............. Equation 5

From the question,

Given: λ = 0.740 m, f = 42 Hz, m = 0.125 kg, L = 2.9 m

Substitute these values into equation 5

T = (42×0.74)²(0.125/2.9)

T = 41.64 N

6 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8 m/s^2. The acce

jeka94

Answer:

The maximum height will be 7408.8 meters

Explanation:

final velocity = initial velocity + acceleration × time

final velocity = 0 m/s + 58.8 m/s^2 ×6 s

Final velocity = 352.8 m/s

final velocity ^2 = initial velocity ^2 + 2 × acceleration × displacement

(352.8)^2 = (0)^2 + 2×58.8 ×displacement

Solving for displacement,

height = 1058.4 meters.

After this, the rocket is in free fall, we can use the same equation.

final velocity ^2 = initial velocity ^2 + 2 ×acceleration×displacement

final velocity = 0

0^2 = 352.8^2 + 2×(-9.8)×displacement

displacement = 6350.4 meters

the maximum height will be 7408.8 meters

3 0

4 years ago

A water pump draws about 9 A when connected to 240 V. What is the cost (with electrical energy at 13 cents per kWh) of running t

Fiesta28 [93]

Answer:

4.49 dollars

Explanation:

i = 9 A, V = 240 V, t = 16 h

Energy = V x i x t = 240 x 9 x 16 = 34560 W h = 34.56 kWh

The cost of 1 kWh is 13 cents.

Cost of 34.56 kWh = 13 x 34.56 = 449.28 cents = 449.28 / 100 = 4.49 dollars

8 0

3 years ago