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zubka84 [21]
3 years ago
8

A speedy tortoise can run with a speed of

bsmiddle" class="latex-formula"> and a hare can run with a speed v_H. The hare waits to rest for a time Δt₀ after the race begins and the tortoise starts and then runs as fast as he can, but the tortoise still wins by a shell (length of shell = s). The length of the race track is a distance, d. The race is considered over when the tortoise crosses the finish line.
Using the symbols of the problem, write an equation for the amount of time that the hare ran, \Delta t_H.
Physics
1 answer:
Vinil7 [7]3 years ago
8 0

Answer:

t_{H}  = t_{T}  - 60

Explanation:

Thinking process:

Let:

distance = speed × time

for the hare:

d_{H}  = s_{H} t_{H}

for the tortoise:

d_{T}  = s_{T} t_{T}

Let the length of the shell be s. Then the two are related. So:

d_{H}  = d_{T}  - s

Let's say the hare ran at x times the speed of the tortoise:

s_{H}  = xs_{T}

So, the equation still holds : t_{H}  = t_{T}  - 60

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I hope this helps you..

3 0
2 years ago
You place the spring vertically with one end on the floor. You then lay a 1.60 kg book on top of the spring and release the book
malfutka [58]

Answer:

Compression in the spring, x = 3.7 cm

Explanation:

It is given that,

Mass of the book, m = 1.6 kg

It can be assumed the spring constant of the spring is, k = 840 N/m

As the book moves down, the change in potential energy of the book is converted to spring potential energy of compression. The mathematical expression is as follows :

\dfrac{1}{2}kx^2=mgx

x=\dfrac{2mg}{k}

x=\dfrac{2\times 1.6\ kg\times 9.8\ m/s^2}{840\ N/m}  

x = 0.037 meters

or

x = 3.7 cm

So, the spring is compressed to a distance of 3.7 cm. Hence, this is the required solution.

4 0
3 years ago
A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

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