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Alex Ar [27]
3 years ago
8

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by θ = 0.220e3t, where

θ is in radians and t is in seconds. Consider a point on the object that is 2.00 cm from the axis of rotation. At t = 0, what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?
Physics
1 answer:
OLga [1]3 years ago
5 0

Answer:a_{t}=3.96

a_{c}=0.8712

Explanation:

Given

\theta =0.220e^{3t}

r=2cm

Now angular velocity is given by \omega =\frac{\mathrm{d}\theta}{\mathrm{d}t}

\omega =0.66e^{3t}

Now linear velocity(v) is given =\omega r

v=1.32e^{3t}

Now tangential component of acceleration is given by

a_{t}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=3.96e^{3t}

at t=0

a_{t}=3.96cm/s^2

radial component of acceleration is given by

a_{c}=\omega ^{2}r

a_{c}=0.4356e^{6t}\times 2

a_{c}=0.8712e^{6t} cm/s^{2}

at t=0

a_c=0.8712 cm/s^{2}

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Answer:

(d) Water has a high specific heat.

Explanation:

At night, when the temperature of earth goes down due to loss of heat , the temperature of water is lost slowly and temperature of land is lost fast because of high specific heat of water . Water loses as well as gains temperature comparatively slowly due to its high specific heat .

During daytime when earth gains heat , the temperature of land rises more rapidly than water so water appears cool even during daytime when land becomes  hotter . It is also due to high heat holding capacity of water or due to high specific heat of water .

8 0
3 years ago
For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul
loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

\frac{P_{o} }{P_{1} } =36.73 (M₁ = 3)

If Po₁ = Po₂

\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825

Table A-1:

\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0
3 years ago
Friction helps your vehicle stop quickly.<br> A. TRUE<br> B. FALSE
meriva

Answer:

true

Explanation:

4 0
2 years ago
Read 2 more answers
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

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a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0
3 years ago
What is rotation in your own words. and you have to be detailed
attashe74 [19]

Answer:

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PS: A middle schooler answered this so if you don't wanna believe me or think I'm wrong because I am younger you do you.

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