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4 years ago

11

Olivia bought new gym shoes to play volleyball because she kept slipping when she ran in her old shoes .how will the new soles h

elp solve the problem

2 answers:

konstantin123 [22]4 years ago

4 0

Answer:

The correct answer is by adding surface roughness to increase friction.

Explanation:

Scrat [10]4 years ago

3 0

By the grip on the bottom

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A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu

ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

$P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}$

Where:

$\gamma = \frac{c_{p}}{c_{v}}$

$\gamma = 1 + \frac{R}{c_{v}}$

Final pressure is:

$P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}} \right)^{\gamma}$

$P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}$

$P_{3} = \frac{P_{o}}{2^{\gamma}}$

8 0

3 years ago

The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the

Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

5 0

3 years ago

In a catapult, energy is transferred into useful kinetic energy stores. What store is the energy in when it is the catapult?

yawa3891 [41]

Answer:

actually I was just wondering what you are thinking

6 0

3 years ago

A graph here shows the displacement of two walkers over time. Waliper 1 is graphed in red while walker 2 is graphed

natita [175]

894984+894390548-098548948+8-734897389696563418686528687

thats your answe thank oyu for asking

5 0

3 years ago

Why could the concentration of matter at the center of an active galaxy like M87 not be made of stars?

romanna [79]

Answer:

an active galaxy is the galaxy that produces large amounts of energy in a very small volume of space.

Explanation:

active galaxies such as the M87 gives out extremely bright straight jets having speeds close to the speed of light providing the evidence that power in active galaxies is concentrated in a small region. this power from such a small region cannot be generated throught nuclear fusion as it happens in stars, this is the region why the concentration of matter in the center of this galaxies is not made up of stars.

6 0

4 years ago