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3 years ago

Which best describes friction in the context of newton's laws?

Physics

2 answers:

Pie3 years ago

5 0

Answer:

Explanation:

friction needs two objects

the two objects need to touch each other

the two objects need to either have relative movement or, tend to have relative movement urge. (kinetic friction or static friction)

kykrilka [37]3 years ago

3 0

Answer:

A force that opposes the relative motion of two objects in contact and sliding past each other.

Explanation:

Friction occurs due to roughness, roughness, or slight surface bumps when two (or more) surfaces are in contact with each other. According to Newton's laws when we push (or pull) an object, it does not move. This is because on this object there is a resistance that the bodies in contact offer to the movement. This counter force is called the friction force F = µ.N (µ coefficient of friction).

That is, for Newton, friction occurs when a force that opposes the relative movement of two objects in contact and sliding one after another.

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A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

Which of the following is always true about a synthesis reaction?

USPshnik [31]

A synthesis reaction forms one product from two reactants.
It has the general form A + X ⇒ AX.

7 0

2 years ago

Students can take the aspire test in ninth and grade

Simora [160]

The answer is Ninth and Tenth grade so the answer would be B

I hope this helps you

3 0

2 years ago

Read 2 more answers

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?

mote1985 [20]

While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

(a) The speed of your phone after 4 s is:

V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

(b) The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

(c) If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

5 0

3 years ago