Ask question

Ask question

All categories

3 years ago

9

Find the magnitude of the gravitational force a 68.4 kg person would experience while standing on the surface of earth with a ma

ss of 5.98 Ã 1024 kg and a radius of 6.37 Ã 106 m. the universal gravitational constant is 6.673 Ã 10â11 n Â· m2 /kg2 . answer in units of n.

1 answer:

luda_lava [24]3 years ago

6 0

Given: Mass of a person M = 68.4 Kg.

Mass of earth Me = 5.98 x 10²⁴ Kg

Radius of earth r = 6.37 x 10⁶ m

G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: F = ?

Formula: F = GMeM/r²

F = (6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴)(68.4 Kg)/(6.37 x 10⁶ m)²

F = 672.41 N

You might be interested in

a slender rod of mass m and length l is released from rest in a horizontal position. what is the rod's angular velocity when it

Makovka662 [10]

Answer:

M g H / 2 = M g L / 2 initial potential energy of rod

I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod

M g L / 2 = 1/3 M L^2 * ω^2 / 2

g = 3 L ω^2

ω = (g / (3 L))^1/2

8 0

2 years ago

A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat

UNO [17]

<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

Force = Spring constant x Elongation

F = kx

Here force is weight of mass

F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x = 36 cm = 0.36 m

Substituting

F = kx

5.3 = k x 0.36

k = 14.72 N/m

Spring constant is 14.72 N/m

6 0

3 years ago

Calculate the average linear momentum of a particle described by the following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2 ,

Maksim231197 [3]

Answer:

a) p=0, b) p=0, c) p= ∞

Explanation:

In quantum mechanics the moment operator is given by

p = - i h’ d φ / dx

h’= h / 2π

We apply this equation to the given wave functions

a) φ = $e^{ikx}$

.d φ dx = i k $e^{ikx}$

We replace

p = h’ k $e^{ikx}$

i i = -1

The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero

p = 0

b) φ = cos kx

p = h’ k sen kx

The average sine function is zero,

p = 0

c) φ = $e^{-ax^{2} }$

d φ / dx = -a 2x $e^{-ax^{2} }$

.p = i a g ’2x $e^{-ax^{2} }$

The average moment is

p = (p₂ + p₁) / 2

p = i a h ’(-∞ + ∞)

p = ∞

6 0

3 years ago

1. Straight-in spaces can leave you a safer out if

Juliette [100K]

Answer:

C. you're able to reverse out of the parking spot

Explanation:

Straight-in parking is an approach of parking that allows a more flexible traffic layout where a driver can approach the spot from either direction and still safely park within the lines. It thus helps to prevent blockage of cars. Each car can move in and out freely preventing it from congestion.

This way of parking can leave you safe when you able to reverse out of the parking spot. It gives you greater control and makes it easier to maneuver out space. The benefits of Straight-in parking are,

- Allows for two-way traffic

- Drivers can line up the vehicle from multiple angles

- Saves time for drivers

3 0

3 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago