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Tju [1.3M]
3 years ago
8

What force on a 936kg elevator that is falling freely at 9.8 m/s^2​

Physics
1 answer:
nataly862011 [7]3 years ago
5 0

Answer:

9172.8 N

Explanation:

F = ma

F = (936 kg)(9.8 m/s^2)

F = 9172.8 N

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What is the acceleration of the 698 kg car that experiences a force of 2410 N?
Nostrana [21]

Hey there!

If you're working with Newton's  Laws / Acceleration, the formula would mostly like look like this

F= MA or you could say → A = \frac{F}{M}

While the F variable represents net    forces of your car  and the m represents the mass of the car mass

Denominator: 2,410

Numerator: 698

Your equation: \frac{2410}{698} ( if you simplify that ←) then you have found your result!

Answer: \boxed{C) 3.45 m/s^2}

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)


7 0
3 years ago
Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and
Lelu [443]

Answer:

The maximum shear stress in shaft AB, T_{ABmax} is 15 MPa

The maximum shear stress in shaft CD,  T_{CDmax} is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = \pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}

Therefore, we have

J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}

Where:

D_{AB} = Diameter of shaft AB = 30 mm = 0.03 m

r_{AB} = Radius of shaft AB = 15 mm = 0.015 m

∴ J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2} = 7.95 × 10⁻⁸ m⁴

and

J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}

Where:

D_{CD} = Diameter of shaft CD = 36 mm = 0.036 m

r_{CD} = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2} = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= 1.58 \times \frac{2\pi }{360} rad = 2.76 × 10⁻² rad.

That is \phi_r = 2.76 × 10⁻² rad.

However  \phi_r =  \phi_{C/D} -  \phi_{B/A}  

Where:

\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G} and

\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}

T_{AB} and T_{CD}= Torque on shaft AB and CD respectively

T_{AB}  = Required

T_{CD}= 500 N·m

L_{AB} and L_{CD} = Length of shafts AB an CD respectively

L_{AB}  = 600 mm = 0.6 m

L_{CD} = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

\phi_r =  \phi_{C/D} -  \phi_{B/A}  =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

2.76 × 10⁻² rad =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

=\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}

Therefore;

T_{AB} =  79.54 N.m

Where T = T_{AB} + T_{CD} =

Therefore T_{CD total } = 500 - 79.54 = 420.46 N·m

τ_{max} = \frac{T\times R}{J}

\tau_{ABmax} = \frac{T_{AB}\times R_{AB}}{J_{AB}} =  \frac{79.54\times 0.015}{7.95\times 10^{-8}} = 15 MPa

\tau_{CDmax} = \frac{T_{CD}\times R_{CD}}{J_{CD}} = \frac{420.46\times 0.018}{1.65\times 10^{-7}} = 45.9 MPa

7 0
3 years ago
Design an inverting amplifier that has a gain of -47 (this gain is negative). Pick resistor values that you have in the lab kit.
Nostrana [21]

Answer:

R2= 470 ohms R1= 10ohms. Schematic is attached

Explanation:

Av=-(R2/R1)

-47=-(470/10)

-47=-47

7 0
3 years ago
Please help asap asap<br> wdtwygcbhyeegdlu tcgbihjlnkdm;fnbkhjg hv
kicyunya [14]

Answer:

B

Explanation:

Bbbb

6 0
3 years ago
Ethan pushes a wooden box across a carpeted floor. Then he pushes the same box across a smooth marble floor. Why does Ethan find
AysviL [449]

Answer:

A. The box experiences more friction on the carpeted floor

Explanation:

Friction is the force that opposes the motion of an object when it slides along a surface. The magnitude of the friction is given by

F=\mu mg

where m is the mass of the object, g is the acceleration due to gravity, and \mu is the coefficient of friction, which depends on the type of material of the surface: the larger this coefficient, the stronger the friction, the more difficult is to push the box along the surface. Generally, a smooth surface has a lower coefficient of friction, while a rough surface has a larger coefficient of friction.

In this case, Ethan find it easier to push the box on the marble floor, because marble is smoother than the carpet and so friction is weaker, while for the carpeted floor the coefficient of friction is larger and so the friction is stronger, making it more difficult to push the box.

3 0
3 years ago
Read 2 more answers
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