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Grace [21]
3 years ago
8

The horizontal component of acceleration, ay during a projectile motion is usually assumed to be_________ a)-9.81 m/s^2 b)-Zero

c)- Constant d)- 32.2 m/s^2
Engineering
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

b) zero

Explanation:

The horizontal component of acceleration during projectile motion is usually assumed to be zero.Because in projectile motion horizontal component of velocity will remain the constant and we know that rate of change of velocity with time is called acceleration.So when velocity is constant then acceleration will be zero.

In projectile motion ,gravitational acceleration will be in only vertical direction.

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GREYUIT [131]

Income

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2 years ago
The author states that chemical engineering is one of the most difficult and complex aspects of engineering. Why do you think th
marishachu [46]

Answer:

hope this helps

Explanation:

answers:

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8 0
3 years ago
Read 2 more answers
Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio
SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

b)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

c)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0
3 years ago
According to the video, what are examples of systems that Stationary Engineers oversee? Check all that apply. electrical systems
garik1379 [7]

Answer:

electrial systems

fire systems

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air systems

Explanation:

3 0
3 years ago
Read 2 more answers
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
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