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Marta_Voda [28]
2 years ago
5

A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =

4ft/s. Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum B, and (b) the number of revolutions executed by drum B during the 4-s interval. 5) A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine its acceleration (a) as it is fired, and (b) as the last particle of fuel is being consumed.
Physics
1 answer:
Ratling [72]2 years ago
5 0

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

v = \omega R\\a = \alpha R

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft

One revolution is equal to the circumference of the drum. So, total number of revolutions is

x / (2\pi R) = 6/(\pi R)

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2

b) a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2

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The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm
Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0
2 years ago
A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th
Tju [1.3M]

Answer:

a = \frac{2}{3}g

T = \frac{mg}{3}

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

mg - T = ma

also by torque equation we can say

TR = I\alpha

TR = \frac{1}{2}mR^2(\frac{a}{R})

T = \frac{1}{2}ma

Now we have

mg - \frac{1}{2}ma = ma

mg = \frac{3}{2}ma

a = \frac{2}{3}g

Also from above equation the tension force in the string is

T = \frac{1}{2}ma

T = \frac{mg}{3}

7 0
2 years ago
A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(
maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

7 0
2 years ago
Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r  were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U =  -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0
2 years ago
The alpha particles leave visible tracks in the cloud chamber because
julia-pushkina [17]
<h2>Answer: Ionization </h2>

The inner atmosphere of a <u>cloud chamber</u> is composed of an easily ionizable gas, this means that little energy is required to extract an electron from an atom. <u>This gas is maintained in the supercooling state, so that a minimum disturbance is enough to condense it</u> in the same way as the water is frozen.

<h2>Then, when a charged particle with enough energy interacts with this gas, it <u>ionizes</u> it. </h2>

This is how alpha particles are able to ionize some atoms of the gas contained inside the chamber when they cross the cloud chamber.

These ionized atoms increase the surface tension of the gas around it allowing it to immediately congregate and condense, making it easily distinguishable inside the chamber like a <u>small cloud</u>. In this way, it is perfectly observable the path the individual particles have traveled, simply by observing the cloud traces left in the condensed gas.

6 0
2 years ago
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