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Rom4ik [11]
3 years ago
11

If 5000 N of thrust is acting to the left, and 4300 N of drag is acting to the right, what is the magnitude and direction of the

resultant force?
Engineering
1 answer:
Kipish [7]3 years ago
8 0

Answer:

700 N acting to the left.

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Help now please evaluate using the commutative property: 40 (32) (10) (25)
Ilia_Sergeevich [38]

Answer:

#See solution for details.

Explanation:

#The commutative property of multiplication tells us that it doesn't matter in what order you multiply numbers. The formula for this property is a * b = b * a:

40\times32\times10\times 25=320,000\\\\25\times10\times32\times 40=320,000\\\\10\times32\times25\times 40=320,000\\\\32\times25\times40\times 10=320,000

Hence, the product of the four numbers remains 320,000 irrespective of their order.

4 0
3 years ago
On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a
stealth61 [152]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

6 0
3 years ago
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
4 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
Cat is stuck in 10 foot hole how do i get it unstuck
Semenov [28]

Answer:

you fill the hole so there is no more cat

7 0
2 years ago
Read 2 more answers
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