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3 years ago

Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high.

2 answers:

8 0

All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.

So we solve for v:

m*g*h=(1/2)*m*v², masses cancel out,

g*h=(1/2)*v², we multiply by 2,

2*g*h=v² and take the square root to get v

√(2*g*h)=v, we plug in the numbers and get:

v=9.9 m/s.

So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.

ivann1987 [24]3 years ago

8 0

Answer:

So we solve for v:

m*g*h=(1/2)*m*v², masses cancel out,

g*h=(1/2)*v², we multiply by 2,

2*g*h=v² and take the square root to get v

√(2*g*h)=v, we plug in the numbers and get:

v=9.9 m/s.

So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.

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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa

Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

$\frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

a ) when charge remains constant

energy = $\frac{q^2}{2C}$

q is charge and C is capacity

energy stored initially E₁= $\frac{q^2}{2C_1}$

energy stored finally E₂ = $\frac{q^2}{2C_2}$

$\frac{E_1}{E_2} = \frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

$E_2$ = $\frac{1.15}{2.3 } \times E_1$

= $\frac{1.15}{2.3 } \times 8.38$

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

$\frac{E_2}{E_1} = \frac{C_2}{C_1}$

$\frac{E_2}{8.38} = \frac{2.3}{1.15}$

$E_2$ = 16.76 .

8 0

3 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

7 0

3 years ago

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

-BARSIC- [3]

Answer:

4.2s

Explanation:

Given parameters:

Power = 2190W

Mass of box = 1.47 x 10⁴g

distance = 6.34 x 10⁴mm

Unknown:

Time = ?

Solution:

Power is the rate at which work is done;

Mathematically;

Power = $\frac{work done}{time}$

Time = $\frac{work done}{power}$

Work done = weight x height

convert mass to kg;

100g = 1kg;

1.47 x 10⁴g = 14.7kg

convert the height to m;

1000mm = 1m

6.34 x 10⁴mm gives 63.4m

Work done = 14.7 x 9.8 x 63.4 = 9133.4J

Time taken = $\frac{9133.4}{2190}$ = 4.2s

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3 years ago

Please help me!!!!!!!!!!Which is true about the actual mechanical advantage of a machine?

Hoochie [10]

B it increases with greater friction

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3 years ago

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ICE Princess25 [194]

Answer:

Explanation:

<em>The most suitable testable question. in this case, would be that 'are there more home runs during the more humid months of the summer?'</em>

Since the aim of the investigation is to find the relationship between humidity and the number of home runs, measuring the number of home runs during the more humid months in the summer and comparing the data to the number of home runs during the less humid months in the same summer would provide the answer.

<u>Only option D raises a valid question that is relevant to the aim of the investigation.</u>

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3 years ago