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3 years ago

Which statement accurately compares the weight of an object on these two planets

Physics

1 answer:

wel3 years ago

5 0

"An object weighs about two times as much on Jupiter as on Neptune." Hope this helps :)

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Ondas de agua en un plato poco profundo tienen 6

castortr0y [4]

I don’t speak Spanish sorry

6 0

3 years ago

A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.

7nadin3 [17]

Answer: $x=\frac{x_m}{\sqrt{2}}$

Explanation:

Given

initially mass is stretched to $x_m$

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is $\frac{kx_m^2}{2}$

Position at which kinetic Energy is equal to Elastic Potential Energy

$K=\frac{mv^2}{2}$

$U=\frac{kx^2}{2}$

it is given

$k=U$

thus $2U=\frac{kx_m^2}{2}$

$2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}$

$2x^2=x_m^2$

$x=\frac{x_m}{\sqrt{2}}$

3 0

4 years ago

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0

3 years ago

Newton's Law of Cooling says that the rate of cooling of an object is proportional to the difference between its own temperature

ASHA 777 [7]

Answer:

$\frac{dQ}{dt} =-hA\Delta T(t)$

Explanation:Newton.s law of cooling states that the rate of cooling of an object is proportional to the difference between its own temperatures and temperature of its surroundings. Mathematically,

$\frac{dQ}{dt} =-hA [T(t)-T(s)]\\$

$\frac{dQ}{dt} =-hA\Delta T(t)$

where $Q$ is the heat transfer

$h$ is heat transfer coefficient

$A$ is the heat transfer surface area

$T$ is the temperature of the object's surface

$T(s)$ is the temperature of the surroundings

3 0

3 years ago

A particle experiences constant acceleration for 20 seconds

lisabon 2012 [21]

Answer:

The correct answer is D. s₂=4s₁

Explanation:

The distance of a particle is given by:

$s=s_{0}+v_{0} t + \frac{1}{2} a t^{2}$

where

s₀ is the initial position when t=0

v₀ is the initial speed when t=0

a is the constant acceleration

t is the time in seconds

Then, the position s₁ is given by:

$s_{1} = s_{0}+v_{0} t_{1} + \frac{1}{2} a (t_{1})^{2}$

As the particle starts from rest v₀=0 and we consider s₀=0, s₁ will be:

$s_{1}= \frac{1}{2} a (t_{1})^{2}$

Furthermore, the position s₂ is:

$s_{2}= \frac{1}{2} a (t_{2})^{2}$

In this case t₁=10 s and t₂=20 s (10 seconds later). In other words t₂=2t₁.

We replace the value of t₂ in the second equation (s₂):

$s_{2}= \frac{1}{2} a (t_{2})^{2} \\ s_{2}= \frac{1}{2} a (2t_{1})^{2} \\ s_{2} = \frac{1}{2} a 2^{2}(t_{1})^{2} \\ s_{2} = \frac{1}{2} a 4 t_{1}^{2}$

Finally, we divide s₂ by s₁ to get the ratio:

$\frac{s_2}{s_1} =\frac{\frac{1}{2}a4t_{1}^{2} }{\frac{1}{2}at_{1}^{2}} =4$

$\frac{s_2}{s_1}=4 \\ s_{2} = 4 s_{1}$

5 0

3 years ago