Name three ways in which an object can accelerate

If a bullet travels at 593.0 m/s, what is its speed in miles per hour?

Ksenya-84 [330]

We have $1 \; mile = 1609.34\; meters$ . So, $1 \; meter = \frac{1}{1609.34} \;mile$ .

$1 \; hour = 3600 \; s$ . So $1 \; s = \frac{1}{3600} \; hour$ .

Thus we can convert the units of the given quantity.

That is,

$593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\
593\;m/s=1,326.51\;miles/hour$ .

The quantity is converted to the required units.

7 0

3 years ago

Does a large object traveling slowly or a small object going fast have a greater momentum?

nekit [7.7K]

Answer:

It depends on the product of their mass and velocity

Explanation:

The momentum of a moving object is defined as the product of its mass and its moving velocity. So a same object would have larger momentum it's traveling faster, or gets heavier, and vice versa. In our case we don't have the exact value of speed and mass of both objects so we can't say which one has a greater momentum.

4 0

3 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

4 years ago

Read 2 more answers

Which one is the best explanation of energy?

vladimir1956 [14]

The energy of a substance is defined as the capacity to do work.The energy may be possessed by the body in various forms.It may be kinetic,potential,magnetic.electric etc.

a- The energy can be transformed from one form to another form.It is called energy conversion.

c-The power defined as the consumption of energy per unit time.

$Mathematically\ power\ P=\frac{E}t} =\frac{W}{t}$