If the same strength force were exerted on both wheelchairs, which chair would go faster?

Engineering

1 answer:

kaheart [24]1 year ago

6 0

Answer:

Hello!!

If the same strength force is exerted on two objects, but the objects have the different masses, the object with less mass will have a greater change in velocity. ... Forces affect motion because they can change an objects motion.

Explanation:

I hope it's help you ✌️✌️

You might be interested in

In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =

monitta

Answer:

Ψ = 10(y^2) + c

y = 1.067m

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =u/y

u = 20y

Ψ = ∫20ydy

Ψ = 10(y^2) + c

(b)

the flow is half below y = 1.6*(2/3)=1.067 m

this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m

3 0

2 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in

RideAnS [48]

To resolve this problem we have,

$R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm$

$F_{f2}$ is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

$\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa$

After that, we can calculate the flexural strenght for the square cross section using the previously value.

$\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN$