9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?

Answers

Answer 1

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)


Related Questions

15 points!

a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.

Answers

Answer:

[tex]4.12\times 10^{-5}\ J[/tex].

Explanation:

Given that,

Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]

Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

[tex]E=\dfrac{Q^2}{2C}[/tex]

Put all the values,

[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]

So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].

03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
and a frequency is 0.59 HZ. Find the perind and spring constant, the maximum speed and
acceleration of the mass, the speed and acceleration when the displacement is 6 cm, compute the
kinetic and the potential energy when the position is 6 cm​

Answers

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

         w = [tex]\sqrt{\frac{k}{m} }[/tex]

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =[tex]\frac{dx}{dt}[/tex]

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = [tex]\frac{dv}{dt}[/tex]

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.

Answers

The answer would be “B” because humans would need water, protection from radiation so we don’t melt or burn to death lol, and a gaseous atmosphere because we would need oxygen.

A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.

What is a Planet?

An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.

By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.

The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.

To know more about Planet:

https://brainly.com/question/14581221

#SPJ2

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