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3 years ago

3. No simple machine is ideal in practice.Why

Physics

1 answer:

Kobotan [32]3 years ago

4 0

Answer:

beacuase

Explanation:

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A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$

The maximum force is when the car is accelerating so

$Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N$

so the maximum force is the maximum force by the maximum speed

$Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW$

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

4 0

4 years ago

If magma is defined as molten rock material,do you need to melt rocks to form magma?<br>

fomenos

In order for magma to form, wet or dry melting of rocks or minerals must occur. Dry melting occurs when minerals or rocks, with no carbon dioxide or water in them, are heated to a specific temperature.

6 0

3 years ago

The pressure exerted by a liquid column of depth 0.5 m on the base of its

Alik [6]

Answer:

1020.4kg/m³

35m³

Explanation:

Given parameters:

Depth = 0.5m

Pressure exerted = 5000N/m²

Unknown = density of the liquid = ?

Solution:

To find the density of this unknown liquid, we use the expression below:

Pressure of liquid = density x height x acceleration due to gravity

5000 = density x 0.5 x 9.8

Density = 1020.4kg/m³

Density of water = 1000kg/m³

Mass of water = 35000kg

Unknown:

Volume of water = ?

Solution:

The volume of water can be derived from the expression below:

Volume = $\frac{mass}{density}$

Volume = $\frac{35000}{1000}$ = 35m³

4 0

3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

4 years ago

I’M DESPERATE PLZ HELP!! 40 POINTS!!

Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, $F_k$ , required is given by the formula, F = m × a,

From which we have;

$F_k$ = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, $F_k$ = 0.0625 N

c. The coefficient of friction between the ice and puck, $\mu_k$ , is given from the equation for the kinetic friction force as follows;

$F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8$

$\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755$

The coefficient of friction between the ice and puck, $\mu_k$ ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force $F_k = 0.012755 \times 1 \times 9.8 = 0.125 \ N$

The acceleration, a = $F_k$ /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here, $F_k = 0.125 \ N$

The coefficient of friction $\mu_k$ ≈ 0.012755 is constant

5 0

3 years ago

3. No simple machine is ideal in practice.Why​

3. No simple machine is ideal in practice.Why