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3 years ago

10

In most circumstances, the normal force acting on an objectand

1 answer:

Lynna [10]3 years ago

4 0

Answer:

Normal Force is usually perpendicular to the movement and static friction usually means that there is no movement.

Explanation:

The work donde by any force on an object is equal to the displacement of the object multiplied by the component of the force that is in the direction of the displacement.

Normal force is usually perpendicular to the movement, so there is no component in the direction of the displacement. This is why it is zero in most circumstances.

<em>Static</em> friction on the other hand, usually means that there is no movement at all (it's static). It means that there is no displacement between the object and ground (in most cases). If there is no displacement, there is no work.

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Explanation:

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3 years ago

A cylindrical rod of stainless steel is insulated on its exterior surface except for the ends. The steady-state temperature dist

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Explanation:

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3 years ago

From the edge of a cliff, a 0.46 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up

wel

Answer:

Explanation:

Given

mass of Projectile(m)=0.46 kg

Initial Kinetic Energy=1430 J

Maximum upward displacement from Launch point=150 m

$K.E.=\frac{mv^2}{2}$

$1430\times 2=0.46\times v^2$

v=78.85 m/s

and $H_{max}=\frac{v^2\sin ^2\theta }{2g}$

$150=\frac{(78.85)^2\sin ^2\theta }{2\times 9.8}$

$\sin \theta =0.687$

$\theta =43.39^{\circ}$

initial horizontal velocity $(v_x)=v\cos \theta$

$v_x=78.84\times \cos (43.39)=57.29 m/s$

Initial vertical velocity $(v_y)=v\sin \theta$

$v_y=78.85\times \sin (43.39)$

$v_y=54.16 m/s$

(c)vertical velocity at any instant=65 m/s

Since initial vertical velocity is 54.16 m/s

so 65 m/s will be acquired when projectile started falling below cliff

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3 years ago

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

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7 0

3 years ago

Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.

lbvjy [14]

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

8 0

3 years ago