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3 years ago

What is the main difference between an internal cumbustion engine and a external combination engine

1 answer:

5 0

Internal and external combustion engines are two types of heat engines: they convert thermal energy into mechanical energy. The main difference between internal and external combustion engine is that in internal combustion engines, the working fluid burns inside the cylinder, whereas in external combustion engines, combustion takes place outside the cylinder and heat is then transferred to the working fluid.

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A rock is thrown from the top of a 20-m building at an angle of 53Â° above the horizontal. If the horizontal range of the throw

NemiM [27]

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

v = 18.22 m/s

6 0

3 years ago

(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca

gulaghasi [49]

Answer:

(a) $E_{ c} = 112.5 \mu J$

(b) $E'_{ c} = 18 \mu J$

Solution:

According to the question:

Capacitance, C = $1.00\mu F = 1.00\times 10^{- 6} F$

Voltage of the battery, $V_{b} = 15.0 V$

(a)The Energy stored in the Capacitor is given by:

$E_{c} = \frac{1}{2}CV_{b}^{2}$

$E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}$

$E_{ c} = 112.5 \mu J$

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

$E'_{c} = \frac{1}{2}CV'_{b}^{2}$

$E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}$

$E'_{ c} = 18 \mu J$

6 0

3 years ago

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

3 years ago

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

charle [14.2K]

The change in internal energy is only gravitional PE because the tube is being drug up at a constant speed. Since it is at a constant speed, the change in KE is 0. Change in PE = m*g*h = 78 kg * 10 m/s^2 * 30 m = 23400 J Work done on the system is from the force Work = force * distance = 350 N * 120 m = 42000 J So, work added 42000 J to the system, but the rider's energy only increased 23400 J. Therefore, friction took up the difference. Friction is where the thermal energy comes from Q = 42000 J - 23400 J = 18600 J. Therfore, friction generated 18600 J of heat to the surroundings.

7 0

3 years ago

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

τ = 26.58 Nm

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

τ = 18.79 Nm

3 0

2 years ago