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3 years ago

By calculating numerical quantities for a multiparticle system, one can get a concrete sense of the meaning of the relationships

Physics

1 answer:

damaskus [11]3 years ago

3 0

Answer:

Explanation:

Given that,

Stiffness K=460N/m

Extension e=0.37m

Two balls attached to a string

First ball has mass and velocity

M1 =8kg and V1~ =4i +11j +0k

For the second ball

M2 =4kg and V2~ =-3i+10j +0k

a. Momentum of system?

Momentum is given as

P = m1•V1~ +m2•V2~

P= 8(4i +11j +0k) +4(-3i+10j +0k)

P=32i + 88j + 0k -12i +40j + 0k

P= 20i + 128j + 0k kgm/s

2. Magnitude of each velocity is given as

V1²=Vx² +Vy² + Vz²

V1² =4²+11²+0²

V1² =137

V1 =√137

V1=11.705m/s

Also V2

V2²= Vx² +Vy² +Vz²

V2²=(-3)²+(10)² +0²

V2²=109

V2 = √109

V2 =10.44m/s

Ktot= k(trans) +k(rel)

d. K(trans) =½m1•V1²

K(trans) =½×8×11.705²

K(trans) = 548.03J

e. K(rel) =½m2•V2²

K(rel) =½ ×4×10.44²

K(rel) = 217.99J

c. Then, K(tot)=K(trans)+k(rel)

K(tot) =548.03+217.99

K(tot) = 766.02J

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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

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3 years ago

The instantaneous velocity of an object is the blank of the object with a blank

miss Akunina [59]

Instantaneous velocity is the velocity at a specific instant in time. I bet you are taking Honors Physics.

3 0

3 years ago

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

You want to raise the temperature of 22kg of water from 60°C to 95°C. If electricity costs $0.08/kWh, how much would it cost you

Sladkaya [172]

Answer:

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Explanation:

dgdfhdhfffff thdhfghjfj e

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3 years ago

Is ‘velocity’the displacement of an object during specific unit of time?

babunello [35]

Answer:

Yes

Explanation:

The displacement covered by a body in unit time is called velocity.

5 0

3 years ago

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