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2 years ago

12

What is A rounded to the nearest tenth?

1 answer:

aniked [119]2 years ago

7 0

Answer: B

Explanition:

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When drawing a free body diagram, which of the following things would NOT be included

Neko [114]

C. Mass of the object

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2 years ago

Can you explain the three main important reason why we use machine

Maru [420]

Answer:

HERE IS YOUR ANSWER

Explanation:

1. The machines makes pur life easier.

2. It makes us do our works faster.

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8 0

2 years ago

Read 2 more answers

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

Katyanochek1 [597]

Answer:

The force is $F = 1041.7N$

Explanation:

The moment of Inertia I is mathematically evaluated as

$I = MR_A^2$

Substituting $1.9kg$ for M(Mass of the wheel) and $\frac{66cm}{2} * \frac{1m}{100cm} = 0.33m$ for $R_A$ (Radius of wheel)

$I = 1.9 * 0.33^2$

$= 0.207kgm^2$

The torque on the wheel due to net force is mathematically represented as

$\tau = FR_B - F_rR_A$

Substituting 135 N for $F_r$ (Force acting on sprocket), $\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m$ for $R_B$ (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

$\tau = F (0.0435) - 135 (0.33)$

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of $\alpha = 3.70 rad/s^2$ and this torque can also be represented mathematically as

$\tau = \alpha I$

Now equating the two equation for torque

$F (0.0435) - 135 (0.33) = \alpha I$

Making F the subject

$F = \frac{\alpha I + (135*0.33) }{0.0435}$

Substituting values

$F = \frac{(3.70 * 0.207) + (135*0.33)}{0.0435}$

$= 1041.7N$

4 0

3 years ago

Please help me with this 29 points

Irina18 [472]

Answer:

)Give the definition of poverty line as defined by the World Bank.

7 0

2 years ago

A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl

Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

$Q = C_w BH^{3/2}$

$H = (\dfrac{Q}{C_wB})^{2/3}$

$H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}$

$H = 0.371\ m$

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0

3 years ago