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kow [346]
3 years ago
6

An external, frictionless pin only prevents rotation. A. True B. False

Physics
1 answer:
stepan [7]3 years ago
7 0

Answer:

False

Explanation:

The roller displacement at an external, friction less rotation pin is found to be zero as it has a support which is no-yielding but still provides rotation to the object where as at an external pin a shear force is present and the moment is absent which means that at an external pin its not the rotation that is absent but its the moment which comes out to be zero.

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Please can anybody tell me what is this lab equipment​
Umnica [9.8K]

Answer:

hii there

its a test tube holder

Explanation:

hope it helps

5 0
2 years ago
Does specific heat change with mass?
ch4aika [34]
Heat<span> capacity ( C ) </span>does change with mass<span>. However, </span>specific heat<span> is the </span>heat<span>capacity per unit </span>mass<span> ( c=Cm ). Therefore if you double the amount of </span>mass<span> in your system, you've doubled its </span>heat<span> capacity, but you've kept the </span>specific heat<span> the same. ... </span>Specific<span> gravity is another such quantity.</span>
7 0
3 years ago
Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:
Anna35 [415]

Answer:

b.  $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate  = 6%

Borrow at the rate of 5% to get

SF200,000/1.05 = $190,476.19

Convert at the spot rate of $0.48 to get

190,476.19*0.48 = $91,428.57

Invest at the interest rate of 6% to get

91,428.57/1.06 = 96,914.28

Therefore, Parker Company will receive $96,914 in 360 days.

7 0
2 years ago
Interchangeably means​
ryzh [129]

Answer:

Interchangeable is something that can be used in the same manner as something else without any important differences.

or in a more simple way

something that can be exchange

For example when two sweaters look almost the same and you could easily wear either one with a particular skirt, this is an example of when the sweaters can be worn interchangeably.

3 0
2 years ago
Read 2 more answers
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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