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3 years ago

A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the

Physics

1 answer:

zlopas [31]3 years ago

8 0

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

$a=2.5 m/s^2$

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

$d=0+\frac{1}{2}(2.5)(35)^2=1531 m$

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What makes a clinical thermometer suitable for measuring small changes in body temperature? *

Nana76 [90]

Answer: Because of the fine bore of the tube.

Explanation:

Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.

The two most common types of themometric fluids for thermometer are alcohol and mercury.

What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.

The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.

4 0

3 years ago

The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu

solniwko [45]

The frequency of the human ear canal is 2.92 kHz.

Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is

Wavelength=4*Length of the ear canal

As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

$wavelength = 4*2.9*10^{-2} =0.116$

Then the frequency is determined as

f=c/λ=339/0.116=2922 Hz=2.92 kHz.

So, the frequency of the human ear canal is 2.92 kHz.

4 0

2 years ago

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

$\frac{v_o}{r_o} = \frac{v_1}{r_1}$

v1 = $\frac{r_1}{r_o} \ \ v_o$

let's calculate

v₁ = $\frac{1.50}{1.00} \ \ 25.0$

v₁ = 37.5 cm / s

4 0

3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

2 years ago