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2 years ago

You ride your bike for a distance of 30 km. You travel at a speed of 0.75 km/ minute. How many minutes does this take?

Physics

2 answers:

My name is Ann [436]2 years ago

5 0

Answer:

Time taken, t = 40 minutes

Explanation:

Given that,

Distance covered, d = 30 km

Speed of the person, v = 0.75 km/min

Time taken by the person is given by :

$t=\dfrac{d}{v}$

$t=\dfrac{30\ km}{0.75\ km/min}$

t = 40 minutes

So, the time taken by the person is 40 minutes. Hence, this is the required solution.

dolphi86 [110]2 years ago

4 0

(30 km)/(0.75 km/min) = 40 min

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Bowen’s reaction series illustrates relations between:

Finger [1]

C. Temperature, chemical composition and mineral structure

Explanation:

The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.

The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.

The left part is the discontinuous end while the right side is the continuous series.
From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
At a relatively low temperature, minerals with frame work structures begins to form . The magma is more enriched with felsic minerals and late stage crystallization occurs here.

Learn more:

Silicate minerals brainly.com/question/4772323

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3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

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2 years ago

How much work does the electric field do in moving a proton from a point with a potential of +130 V

In-s [12.5K]

Explanation:

iy finds so she said !gWelcome to Gboard clipboard, any text that you copy will be saved here.Welcome to Gboard clipboard, any text that you copy will be saved here.Welcome to Gboard clipboard, any text that you copy will be saved here.Tap on a clip to paste it in the text box.Tap on a clip to paste it in the text box.Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.Use the edit icon to pin, add or delete clips.Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.Use the edit icon to pin, add or delete clips.

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3 years ago

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

pogonyaev

Let h = distance (m) to the water surface.

Initial velocity, u = 0 (because the stone was dropped).
Use the formula
h = ut + (1/2)gt^2
where g = 9.8 m/s^2 (acc. due to graity)
t = time (s)

h = (1/2)*(9.8)*(3^2) = 44.1 m

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3 years ago

A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A i

SOVA2 [1]

Answer:

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 32.4 J

mechanical energy after collision = 32.433 J

Explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle = 28°

to find out

final velocity and mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 = 0.1765 V1 + 0.09389 v2 ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

0 = 0.09389 v1 - 0.1768 v2 .......................2

from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

mechanical energy before collision = 32.4 J

and

mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²

mechanical energy after collision = 32.433 J

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2 years ago