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3 years ago

9As a student walks downhill at constant speed, his gravitational potential energy

Physics

1 answer:

Minchanka [31]3 years ago

4 0

Answer:

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Explanation:

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Emily is a competitive swimmer. Why does wearing a swim cap help Emily swim faster?

sladkih [1.3K]

Answer:

It's helps Emily became when she swims her hair is not in the way creating friction and making her swim faster. friction acts between two metals blocks that slide past each other.

Explanation:

6 0

3 years ago

A block with a mass M = 4.85 kg is resting on a slide that has a curved surface. There is no friction. The speed of the block af

natali 33 [55]

Answer:

The correct option is a

Explanation:

From the question we are told that

The mass of the block is $m = 4.84 \ kg$

The height of the vertical drop is $h = 19.6 \ m$

Generally from the law of energy conservation , the potential energy at the top of the slide is equal to the kinetic energy at the point after sliding this can be mathematically represented as

$PE = KE$

i.e $m * g * h = \frac{1}{2} * m * v^2$

=> $gh = 0.5 v^2$

=> $v = \sqrt{\frac{9.8 * 19.6}{0.5 } }$

=> $v = 19.6 \ m/s$

4 0

2 years ago

Which animal in the rainforest is not nocturnal?

Stels [109]

Answer:

tree frog i believe hope its right

5 0

3 years ago

2. A student drives 7.8-km trip to school and averages a speed of

Alekssandra [29.7K]

Answer:

The total time is: t=451.22 sec

The average speed is: V=34.57 m/s

Explanation:

Average speed

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

$\displaystyle V=\frac{x}{t}$

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

$\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec$

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

$\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec$

The total time is:

$t=239.26\ sec+211.96\ sec=451.22\ sec$

$\boxed{t=451.22\ sec}$

The average speed is:

$\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s$

$\boxed{\displaystyle V=34.57\ m/s}$

6 0

3 years ago

A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy

Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be $F_1$ and force on +Q charge y axis due to -Q charge on x-axis be $F_2$ .

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = $\sqrt{2}d$ units

$k_e$ = Coulomb constant

$F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N$

$F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N$

Net force on +Q charge on y-axis is:

$F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N$

$F_y=F_1-F_2cos45^o$

$F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})$

$F_N=\sqrt{F_x^2+F_y^2}$

$|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|$

The net froce on the +Q charge on y-axis is

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

4 0

3 years ago

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