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erma4kov [3.2K]

2 years ago

14

Describe your acceleration if you ride your bike up a hill,then ride down the other side.

1 answer:

Ghella [55]2 years ago

5 0

Answer:

first negative and then positve

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What is the speed of a horse in meters per second that runs a distance of 1.2 miles in 2.4 minutes

Ilia_Sergeevich [38]

Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]

4 0

2 years ago

Will give brainliest

NNADVOKAT [17]

I love science! if you need any more help let me know i cant guarantee i can help but i will try!

The correct answer is "Some substances must be dissolved in water before they can be used".

5 0

3 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

You throw a ball with a speed of 25.0 m/s at an angle of 40.0â—¦ above the horizontal directly toward a wall. The wall is 22.0 m

kirza4 [7]

The horizontal speed is going to be the cosine of the given speed, therefore, the horizontal speed is 19.15 m/s. To find the time, divide the 22 m distance by the velocity. This results in 1.131 seconds, which is in between C and D.

3 0

3 years ago

Read 2 more answers

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

2)

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

3)

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

4)

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

2 years ago