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3 years ago

Which value is equivalent to 7.2 kilograms?

Physics

1 answer:

emmainna [20.7K]3 years ago

3 0

7200 grams is what equivalen to 7.s kilograms

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A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximu

Mrac [35]

Answer:

x=±0.026m

Explanation:

In simple harmonic motion the maximum value of the magnitude of velocity

$v_{max}=wA=\sqrt{\frac{k}{m} }A$

The speed as a function of position for simple harmonic oscillator is given by

$v=w\sqrt{A^{2}-x^{2}$

where A is amplitude of motion

Given data

Amplitude A=3 cm =0.03 m

v=(1/2)Vmax

To find

We have asked to find position x does its speed equal half of is maximum speed

Solution

The speed of the particle the maximum speed as:

$v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}$

x=±(√3(0.03)/2)

x=±0.026m

5 0

4 years ago

The impulse experienced by a body is equivalent to the body’s change in

nika2105 [10]

<h2>Answer: </h2>

Momentum

<h2>Explanation: </h2>

The momentum of a particle is defined as the product of the particle mass and the particle velocity as follows:

$\overrightarrow{p}=m\overrightarrow{v}$

On the other hand, the impulse of a constant force is defined as:

$\overrightarrow{J}=\varSigma\overrightarrow{F}(t_{2}-t_{1})=\varSigma\overrightarrow{F}\Delta t$

We also know that the net force acting on a particle equals the rate of change of the particle’s momentum, so:

$\varSigma\overrightarrow{F}=m\overrightarrow{a}=m\frac{d}{dt}(\overrightarrow{v})=\frac{d}{dt}(m\overrightarrow{v})=\frac{d\overrightarrow{p}}{dt}$

If the force is constant, then $\frac{d\overrightarrow{p}}{dt}$ equals the total change in momentum over a period of time:

$\varSigma\overrightarrow{F}=\frac{\overrightarrow{p_{2}}-\overrightarrow{p_{1}}}{t_{2}-t_{1}} \\ \\ \varSigma\overrightarrow{F}(t_{2}-t_{1})=\overrightarrow{p_{2}}-\overrightarrow{p_{1}} \\ \\ \boxed{\overrightarrow{J}=\Delta \overrightarrow{p}}$

3 0

3 years ago

The majority of which type of dancing was created in African American communities?

icang [17]

Answer:

THE ANSWER IS THE SWING

Explanation:

4 0

2 years ago

Read 2 more answers

Please help 9.2.1 project in science just ned an example

olga55 [171]

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

3 0

3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$