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2 years ago

Pls help me i will name brainliest

Physics

2 answers:

Mandarinka [93]2 years ago

7 0

Answer:

a for the first one b for the second

jolli1 [7]2 years ago

6 0

Agreed when looking at the law it will end up with a for firs answer and b for the second

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A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago

Which career would be the best for someone who wants to work in entertainment but does not want to be in front of any type of au

GalinKa [24]

directing, your out of the spotlight

8 0

3 years ago

Read 2 more answers

A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement

Igoryamba

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

$\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}$

The angular displacement is given by:

$\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}$

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

$\displaystyle \alpha=\frac{6-13.5}{7}$

$\displaystyle \alpha=\frac{-7.5}{7}$

$\displaystyle \alpha=-1.071 \ rad/s^2$

Thus, the angular displacement is:

$\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}$

$\displaystyle \theta=94.5-26.25$

$\boxed{\displaystyle \theta=68.25\ rad}$

Angular displacement=68.25 rad

3 0

3 years ago

I will give 14 points & make you the brainiest

Virty [35]

Answer:

B. silicate rocks and metals

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3 years ago

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almond37 [142]

City 4 because it is closer to the equator

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3 years ago