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dem82 [27]
3 years ago
14

If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially,

the block is at rest
Engineering
1 answer:
grigory [225]3 years ago
3 0

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

F_{H} = F cos0

F_{H} = (30) x 4/5

F_{H} = 24N

Now Calculate V using following formula

V = V_{0} + at

V_{0} = 0

a = F_{H} / m

a = 24N / 20 kg

a = 1.2m / S^{2}

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = F_{H} x V

Power = 24 x 4.8

Power = 115.2 W

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A ballistic pendulum consists of a 3.60 kg wooden block on the end of a long string. From the pivot point to the center‐of‐mass
Pavel [41]

Answer:

17.799°

Explanation:

When the bullet hits the block at that time the momentum is conserved

So, initial momentum = final momentum

P_i=P_f

So 28\times 10^{-3}\times 210=(3.6+0.028)v_f

v_f=1.6207\ m/sec

Now energy is also conserved

So \frac{1}{2}\times (3.6+0.028)\times 1.6207^2=(3.6+0.028)\times 9.81\times 2.8(1-cos\Theta )

cos\Theta =0.8521So\ \Theta =17.799^{\circ}

3 0
2 years ago
Why does the compression-refrigeration cycle have a high-pressure side and a low-pressure side?
Cloud [144]

Answer: D

Explanation:

8 0
2 years ago
Heats of Reaction and Hess's Law experiment tells you to look up three values. Write down these values. Be sure to clearly indic
Anna [14]

Answer:

Enthalpy of reaction (kJoules/mole)

Heat of formation of products (kJoules/mole)

Heat of reaction of reactants (kJoules/mole)

Explanation:

The general expression for calculating the overall enthalpy of reaction is given as following:

ΔH = ∑ΔH[producst] - ∑Δ[reactants]

Thus, the heat of reaction is given as the difference between the formation of the products and the formation of the reactants. The units are expressed as kJ/mol of reactants or products.

Thus, the three values are fundamental in the determination of the overall energy of the reaction from Hess' Law.

8 0
2 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
2 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
Arada [10]

Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

5 0
3 years ago
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