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3 years ago

10

A object has a mass of 13 Kg . - 17 Calculate the mass of this object on the surface of Jupiter . ( The gravity of Jupiter is 26

N / kg )

1 answer:

Nonamiya [84]3 years ago

7 0

Explanation:

It is given that,

Mass of an object is 13 kg

We need to find the mass of this object on the surface of Jupiter.

The mass of an object is the amount present in it. It it independent of location. It mass remains the same at Jupiter as well. Its does not depend on the gravity on that surface.

Hence, its mass will remain 13 kg on the surface of Jupiter.

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You jump off a platform 134 m above the Neuse River. After you have free-fallen for the first 40 m, the bungee cord attached to

omeli [17]

Answer:

The acceleration at lowest point is 19.62 m/s^2

Explanation:

Conservation of energy is an concept in which it is stated that the energy of an isolated object remains the same. Energy changes from one form to another.

Lets Assume

Constant of string is K

By using the conservation of energy we will have the following equation

1/2 x 80^2 x K = m x 9.81 x 120

3200 K = 1177.2 m

K = 1177.2 m / 3200

K = 0.368 m

At the lowest point we will have

a = ( K x X - m x g ) / m

a = ( 0.368 m x 80 - m x 9.81 ) / m

a = 19.62 m / s^2

So, the acceleration at lowest point is 19.62 m/s^2

7 0

3 years ago

Read 2 more answers

What is the purpose of using conductors in a circuit?

Aleonysh [2.5K]

Answer:

Conductors and insulators are both important in the field of electronics. Electrical conductors allow electric current to flow easily because of the make up of their atoms. In a conductor, the outer electrons of the atom are loosely bound and can freely move through the material when an electric charge is applied.

Explanation:

4 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

The momentum of light, as it is for particles, is exactly reversed when a photon is reflected straight back from a mirror, assum

LenKa [72]

Answer:

a) E = 2.00 10³ J , b) I = 6.66 10⁻⁶ N s , c) F = 1.66 10⁻⁶ N

Explanation:

a) The intensity is defined as the power per unit area

I = P / A

P = I A

Power is energy for time

P = E / t

We replace

E / t = I A

E = I A t

E = 1.0 10³ 2.0 1.00

E = 2.00 10³ J

b) The moment is

p = U / c

In the case of a reflection the speed is reversed, so the moment

Δp = 2 U / c

I = Δp

I = 2 U / c

I = 2.00 10³/3 10⁸

I = 6.66 10⁻⁶ N s

c) The defined impulse is

I = F t

F = I / t

For a time of 1 s

F = 6.66 10⁻⁶ / 1

F = 1.66 10⁻⁶ N

d) Suppose n small mass mirror m = 10 10⁻³ kg, we write Newton's second law

F = ma

a = F / m

a = 1.66 10⁻⁶ / 10 10⁻³

a = 1.66 10⁻⁴ m / s

We see that the acceleration is very small and attended to increase the mass of the mirror will be less and less, so the assumption of no twisting of the mirror is very reasonable

5 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

BartSMP [9]

No correlation because it is neither positive or negative. The dots are scattered so that would be no correlation.

7 0

3 years ago