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2 years ago

13

An electric fan rotates at 800rev/min.consider a point on the blade a distance of 0.16m from the axis, calculate the velocity of

this point and it's centripetal acceleration

1 answer:

Anuta_ua [19.1K]2 years ago

6 0

Answer:

A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m.

Explanation:

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If the price of gasoline at a particular station in Europe is 5 euros per liter. An American student in Europe is allowed to use

Mashcka [7]

Answer:

Number of gallons =2 gallon

Explanation:

given data:

rate of gasoline ineurope = 5 euro per liter

total money to buy gasoline = 40 euro

total gasoline an american can buy in europe $= \frac{40}{5}$

= 8 litres of gasoline

As given in the question 1 ltr is 1 quarts therefore

Total no. of quarts is 8 quarts

As from question 4 quarts is equal to one gallon, hence

Number of gallons $= \frac{8}{4} = 2 gallon$

5 0

3 years ago

M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.

MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is $6.49\times10^6m/s$

<h3>How to calculate the speed of the electron?</h3>

We know, that the energy of the system is always conserved.

Using the Law of Conservation of energy,

$\triangle K+\triangle$ U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

$(\frac{1}{2} mv^2)+(-q\triangle V)$ =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let $v_f$ and $v_i$ represent the final and initial speed.

Here, $v_i$ =0

Solving for $v_f$ , we get:

$v_f=\sqrt{\frac{-2q\triangle V}{m} }$

$=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }$

=6.49 $\times10^6$ m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

4 0

1 year ago

You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine

DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

$F_x = F cos \theta$

with $\theta=41.9^{\circ}$

- The frictional force, whose magnitude is

$F_f = \mu mg$

where $\mu=0.33$ , m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

$F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N$

8 0

3 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the je

jenyasd209 [6]

<h2>The answer got is reasonable.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s

Acceleration, a = ?

Final velocity, v = 400 m/s

Displacement,s = 4 km = 4000 m

Substituting

v² = u² + 2as

400² = 300² + 2 x a x 4000

a = 8.75 m/s² = 8.8 m/s²

The acceleration is 8.8 m/s²

The answer got is reasonable.

7 0

3 years ago