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antiseptic1488 [7]

3 years ago

14

A car starts from rest and accelerates uniformly at 6.6 m/s2 for 10.s. How far does the car travel?

1 answer:

quester [9]3 years ago

5 0

Explanation:

s = ut + 1/2at²

= 0(10) + 1/2×6.6×(10)²

= 3.3 × 100

= 330 m

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How much force is required to accelerate a 22Kg mass at 6 m/s?

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A system consists of two spheres, of mass m and 2m, connected by a rod of negligible mass, as shown above. The system is held at

Svetradugi [14.3K]

Answer:

Rod will remain horizontal all the time after release.

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This is because net torque on the rod about any point in space is zero.

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8 0

3 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

4 years ago