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3 years ago

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A skier has an acceleration of 2.5 m/s2. How long does it take her to come to a complete stop from a speed of 18 m/s? A. 3.4 s B

. 7.2 s C. 15.5 s D. 0.14 s

1 answer:

noname [10]3 years ago

6 0

The question is defective. If her acceleration is 2.5 m/s/s, then her speed is 2.5 m FASTER each second, and she will never stop.

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Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i

Serggg [28]

Answer:

2.2N

Explanation:

Given parameters:

Work done = 379.5J

Height = 173m

Unknown:

Amount of force exerted on the sled = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

Work done = mgh

m is the mass

g is the acceleration due to gravity

h is the height

mg = weight;

Work done = weight x h

379.5 = weight x 173

weight = $\frac{379.5}{173}$ = 2.2N

4 0

2 years ago

Two power lines run parallel for a distance of 220 m and are separated by a distance of 40.0 cm. If the current in each of the t

daser333 [38]

Answer:

The magnitude of force is 1.86 N and the direction of force is towards the other wire.

Explanation:

Given:

Current flowing through each power line, I = 130 A

Distance between the two power lines, d = 40 cm = 0.4 m

Length of power lines, L = 220 m

The force exerted by the power lines on each other is given by the relation:

$F = \frac{\mu_{0}LII }{2\pi d}$

Substitute the suitable values in the above equation.

$F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}$

F = 1.86 N

Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.

5 0

3 years ago

A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago

What property or properties of a wave determine it's speed?

mash [69]

The properties of the wave don't determine its speed. The properties of the medium do. You can FIND the speed by measuring the wave's frequency and wavelength.

7 0

3 years ago

If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?

Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

$A=100 mm^2$

multiplied by the height of rain that fell, which is

$h=5.0 mm$

Therefore, the volume is

$V=hA=(5 mm)(100 mm^2)=500 mm^3$

7 0

3 years ago