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creativ13 [48]
3 years ago
13

Name the physical quantity which changes contenously during uniform-circular motion.

Physics
1 answer:
kumpel [21]3 years ago
8 0
The direction of the motion is constantly changing during
motion over any closed path, not only circular.
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Why did the Founding Fathers want to include a bill of rights in the U.S.
sergey [27]

Answer:

Its A

Explanation:

Thats because federalists wanted to take away rights. But the fathers wanted to definitely ensure that each person had equal rights and liberty.

7 0
3 years ago
A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top
Sveta_85 [38]

Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.

Given,

Length, L = 4 m

Outer diameter, D = 300mm, D= 0.3 m

Thickness, t = 50 mm, t = 0.05 m

Stress produced, σ = 75000 kN/m²

Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²

Calculating the diameter of the cylinder,

Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)

d= 0.2 m

(i) Magnitude of the load P:

Using the relation, σ =P/A

P = σ × A = 75000 × π /4 (D² – d² )

P= 75000 × π/4 (0.3² – 0.2²)

P= 75000 × π/4 (0.09 – 0.04)

P = 2945.2 kN

Hence, Magnitude of the load P is 2945.2 kN.

(ii) Longitudinal strain produced, e :

Using the relation, Strain, (e) = stress/E

e= 75000/(1.5 x 10⁸)= 0.0005

Hence, the Longitudinal strain produced is 0.0005.

(iii)Total decrease in length, dL:

The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.

Strain = change in length/original length

e= dL/L

0.0005 = dL/4

dL = 0.0005 × 4m = 0.002m=2mm

Hence,the decrease in length is 2 mm.

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5 0
2 years ago
A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s
skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I =  0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0
3 years ago
Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi
insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0
3 years ago
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If you can’t open the lid of a jelly jar what should you do?
9966 [12]

If I can't open the lid of a jelly jar, I'd keep trying and if I can't open the lid of a jelly jar after the MANY tries I took, I'd ask for help.

6 0
3 years ago
Read 2 more answers
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