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3 years ago

1. How much force is needed to accelerate a 66 kg skier at 2 m/sec2?

Physics

1 answer:

Tju [1.3M]3 years ago

3 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 66 × 2

We have the final answer as

Hope this helps you

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In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nu

nikitadnepr [17]

Answer:

Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)

Kinetic energy at the beginning: KEi = $mv^{2}$ /2

Potential energy at the end: Uf =(Qq/r)*k

Kinetic energy at the beginning: KEf = 0

From energy conservation:

r = 2*Qk/ $v^{2}$ *q/m

Ranking of the closest approach to the gold nucleous comes from q/m:

- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest

- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking

- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4

Explanation:

Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)

Kinetic energy at the beginning: KEi = $mv^{2}$ /2

Potential energy at the end: Uf =(Qq/r)*k

Kinetic energy at the beginning: KEf = 0

From energy conservation:

r = 2*Qk/ $v^{2}$ *q/m

Ranking of the closest approach to the gold nucleous comes from q/m:

- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest

- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking

- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4

7 0

3 years ago

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

$K_{1} = K_{2} + W_{loss}$

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energy, measured in joules.

$W_{loss}$ - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s$

Where:

$v_{1}$ , $v_{2}$ - Initial and final speed of the skater, measured in meters per second.

$m$ - Mass of the skater, measured in kilograms.

$\Delta s$ - Length of the rough patch, measured in meters.

$f$ - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

$f = r \cdot m \cdot g$

Where:

$r$ - Ratio of friction force to weight, dimensionless.

$g$ - Gravitational constant, measured in meters per square second.

Then,

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s$

The equation is simplified algebraically and patch length is cleared afterwards:

$\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s$

$\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }$

Given that $v_{1} = 5\,\frac{m}{s}$ , $v_{2} = 2.5\,\frac{m}{s}$ , $r = 0.22$ and $g = 9.807 \,\frac{m}{s^{2}}$ , the length of the rough patch is:

$\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta s = 4.345\,m$

The length of the rough patch is 4.345 meters.

6 0

3 years ago

Find the speed vfinal of the joined cars after the collision. mastering physics

Tanya [424]

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6 0

4 years ago

Anyone know this... pls help its for a grade

hodyreva [135]

His average speed is 45 miles an hour

3 0

3 years ago

A piano has a mass of 99 kg. What is the weight of the piano?

Blababa [14]

Explanation:

weight of the piano = mg

w = 99 x 10 =990 N

7 0

3 years ago