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3 years ago

9

Mrs. Botts applies brakes on a car help it to decelerate at the rate of -0.80m/s2. What distance is required to stop the car whe

n it is moving 17 m/s.

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

Explanation:

vf=0

vi = 17

a = -0.8

Δx = ?

vf^2 = vi^2 + 2 a Δx

Δx = (vf^2 - vi^2) / 2a

Δx = (0-17^2) / 2 (-0.8)

Δx = 180.625 m

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To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially

cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside a pipe between two points.

Thus

⇒ $A_{1}$ × $V_{1}$ = $A_{2}$ × $V_{2}$

⇒ $\frac{\pi }{4}$ × $d_{1} ^{2}$ × $V_{1}$ =

⇒ $d_{1} ^{2}$ × $V_{1}$ = $d_{2} ^{2}$ × $V_{2}$

⇒ $(3.2)^{2}$ × 110 = $(0.57)^{2}$ × $V_{2}$

⇒ $V_{2}$ = 3467 cm / sec

⇒ $V_{2}$ = 34.67 m / sec

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0

3 years ago

Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri

Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

(c) 0.716 s

Explanation:

Gravity would be exerting on the 2 masses

$G_1 = Mg = 15.3*9.81 = 150.093N$

$G_2 = mg = 8.29*9.81= 81.32N$

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

$F = G_1 - G_2 = 68.77N$

This force would generate torque on the solid pulley

$T = FR = 68.77 * 0.0075 = 0.5158 Nm$

We can also calculate the pulley moments of inertia, with it being solid

$I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2$

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

$\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2$

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

$\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad$

(c)The time it takes to get to this distance is

$\theta = \frac{\alpha t^2}{2}$

$t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513$

$t = \sqrt{0.513} = 0.716s$

(b)The final angular speed of the disk is

$\omega = \alpha t = 1300.58*0.716 = 931 rad/s$

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

$v = \omega R = 931*0.0075 \approx 7m/s$

5 0

3 years ago

Which of the device has the highest resistance?

-BARSIC- [3]

It would be c) Rheostat

5 0

3 years ago

Read 2 more answers

A paper filled capacitor is charged to a potential difference of 2.1 V and then disconnected from the charging circuit. The diel

Goryan [66]

Answer:

Part a)

$V' = 7.77 Volts$

Part b)

$k' = 6.27$

Explanation:

As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery

So here we can say that charge on the plates will remain conserved

So we will have

$Q = kC(2.1)$

now dielectric is removed between the plates of capacitor

so new potential difference between the plates

$V' = \frac{Q}{C'}$

$V' = \frac{kC(2.1)}{C}$

$V' = 3.7 \times 2.1$

$V' = 7.77 Volts$

Part b)

Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates

So again charge is same so potential difference is given as

$V" = \frac{Q}{k'C}$

$0.59 V = \frac{kCV}{k'C}$

$0.59 = \frac{3.7}{k'}$

$k' = 6.27$

5 0

3 years ago

An object is projected straight upward with an initial speed of 25.0 m/s

hoa [83]

Answer:

What's the question your asking?

Explanation:

It moves with an initial speed of 25.0 miles per second.

4 0

3 years ago