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strojnjashka [21]
3 years ago
13

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ed cm/s squared? what was the acceleration of the cart with high fan speed cm/s squared?

Physics
2 answers:
ArbitrLikvidat [17]3 years ago
8 0

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

mixer [17]3 years ago
5 0

Answer:

Explanation:

Low fan speed: 18.0 cm/s2

Medium fan speed: 24.0 cm/s2

High fan speed: 32.0 cm/s2

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Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

\dfrac{dF_{net}}{df}=0

We need to calculate the unknown force

Using formula of net force

\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}

Put the value into the formula

\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}

\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}

The magnitude of net force,

F_{net}=\sqrt{F_{x}^2+F_{y}^2}

F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}

F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}

F_{net}=\sqrt{F^2+4899.78+36.232F}

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(\dfrac{dF_{net}}{dF})^2=2F+36.232

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3 years ago
Help pleasseeee URGENT
Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

When a collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum is simplified to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}

\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}

\displaystyle v'_1=\frac{0.34}{0.16}

v'_1=2.125\ m/s

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Explanation:

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