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3 years ago

Need ASAP!! An object lies motionless on a rough inclined surface .

Physics

2 answers:

PSYCHO15rus [73]3 years ago

7 0

Answer:

yes

Explanation:

yes

Aleks [24]3 years ago

5 0

Answer:

that's true

Explanation:

if the rough inclined plane was rough enough than it would be true

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A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

6 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Which of the following statements must be true?

anyanavicka [17]

Answer:

your answer is B. The velocity could be in any direction, but the acceleration is in the direction of the resultant force

6 0

4 years ago

A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless.

aleksandrvk [35]

Answer:

The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi

. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I

Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum

Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi

8 0

3 years ago

What’s the most distant object in the solar system

Scrat [10]

Answer:

pluto

Explanation:

i had the same question

7 0

3 years ago

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