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3 years ago

What is the mass of the rock?

Physics

1 answer:

kow [346]3 years ago

5 0

Answer:

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4.6 i think

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The magnitude of the momentum of an object is 64 kilogram meters per second. If the magnitude of te velocity is doubled, the mag

dsp73

Answer:

$128 \frac{kg*m}{s}$

Explanation:

$P=m*v\\if double velocity:\\P=m*2v=2m*v\\$

7 0

4 years ago

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

1. A change in which one or more substances are converted into new substances is a

Ratling [72]

Answer:

<u>CHEMICAL CHANGE</u>:

A change in which one or more substances are converted into new substances is a <em>chemical change</em>.

<u>EXPLANATION:</u>

Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances.

<u>EXAMPLE:</u>

<em>Examples of Chemical Change in Everyday Life </em>

Burning of paper and log of wood.

Digestion of food.

Boiling an egg.

Chemical battery usage.

Electroplating a metal.

Baking a cake.

Milk going sour.

Various metabolic reactions that take place in the cells.

4 0

3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$