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Alina [70]
3 years ago
6

you are at the grand canyon standing at the edge of a ledge 1857 m high you have a mass of 61 kg. you decide to take a selfie to

share with your science teacher when you get home which causes you to wonder how much gravitational potential energy do you have standing at the edge of this cliff
Physics
1 answer:
evablogger [386]3 years ago
7 0

Answer:

1,110,114.6 J

Explanation:

The height is 1857m, the mass is 61 kg, and the accelaration is 9.8 m/s^2. Plug in the equation GPE=h*m*a and you get 1,110,114.6.

h= height

m= mass

a= accelaration

GPE= gravitational potential energy

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A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction
Nuetrik [128]

Answer:

brainly.com/question/5055678

Explanation:

4 0
3 years ago
What piece of sports equipment is aerodynamic
Mashcka [7]
Aerobie.  Frisbee.  Discus.  Javelin.  I suppose an American football to some extent. 
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6 0
3 years ago
The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e
FrozenT [24]

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

8 0
3 years ago
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
Flauer [41]

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 = \sqrt{\frac{k}{m} }

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 =\frac{dx}{dt}

          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 = \frac{dv}{dt}

            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

5 0
3 years ago
You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do
melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

W = F_g .d

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0
3 years ago
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