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3 years ago

The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3

Physics

1 answer:

Sav [38]3 years ago

6 0

Answer:

$a$ has units of distance

$b$ has units of distance over time

$c$ has units of distance over $time^2$

$d$ has units of distance over $time^3$

Explanation:

Since the expression for the distance is:

$x = a+b\,t+c\,t^2+d\,t^3$

then:

$a$ has units of distance

$b$ has units of distance over time

$c$ has units of distance over $time^2$

$d$ has units of distance over $time^3$

because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.

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An object start from rest with a constant acceleration of 8.00 m/s^2 along straight line.

san4es73 [151]

I’ll say c Bc it make more since to find the travel distance

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3 years ago

After asking a question, a scientist can form a(n) , which is an idea that may be proved or disproved by an experiment.

DochEvi [55]

Answer:

When scientists have a question, they form a hypothesis, which is an idea that may be proved or disproved by an experiment.

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3 years ago

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A ball that is thrown upwards from the ground will eventually reach its highest point and fall back to the ground. Which of the

tia_tia [17]

Answer:

option (c)

Explanation:

When an object thrown upwards, the value of acceleration acting on the object is acceleration due to gravity which is always acting towards the earth.

As it falls downwards, the acceleration is again equal to the acceleration due to gravity.

So, the ball's acceleration is constant.

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3 years ago

What is the magnitude of the gravitational force acting on the earth due to the sun?

expeople1 [14]

Answer: $3.524(10)^{22}N$

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m1=1.99(10)^{30}kg$ is the mass of the Sun

$m2=5.972(10)^{24}kg$ is the mass of the Earth

$r=1.50(10)^{11}m$ is the distance between the Sun and the Earth

Substituting the values in (1):

$F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}$ (2)

Finally:

$F=3.524(10)^{22}N$ This is the gravitational force acting on the earth due to the sun

3 0

3 years ago

Read 2 more answers

*PHYSICS HELP*

sveta [45]

My calculator is about 1cm thick, 7cm wide, and 13cm long.

Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .

The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.

If its density is actually 1 g/cm³, then its mass is about 91 grams.

The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.

My result of 91 grams is about 100 grams ... about 10² grams.

Your results could be different.

3 0

2 years ago