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A snowball that will be used to build a snowman is at the top of a only hill. If the

Darina [25.2K]

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

$E=mgh$

g is acceleration due to gravity

h is height of the hill

$h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m$

So, the height of the hill is 24.11 m.

5 0

3 years ago

The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a

SpyIntel [72]

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

$a=\dfrac{dv}{dt}$

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, $a=\dfrac{-60\ m/s}{10\ s}$

$a = -6\ m/s^2$

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. $-6\ m/s^2$ .

6 0

3 years ago

. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

(c) 16 m/s²

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

3 years ago

If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52

castortr0y [4]

Answer:

$A.\hspace{3}n=1.52$

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

$n=\frac{c}{v}$ (1)

Where:

$c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\$

$v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium$

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

$122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}$

Finally, replacing the data in (1):

$n=\frac{300,000}{196339.48} =1.527965746\approx1.52$

3 0

3 years ago

How many normal modes of oscillation or natural frequencies does each if the following have: (

Vadim26 [7]

Each of these systems has exactly one degree of freedom and hence only one natural frequency obtained by solving the differential equation describing the respective motions. For the case of the simple pendulum of length L the governing differential equation is d^2x/dt^2 = - gx/L with the natural frequency f = 1/(2π) √(g/L). For the mass-spring system the governing differential equation is m d^2x/dt^2 = - kx (k is the spring constant) with the natural frequency ω = √(k/m). Note that the normal modes are also called resonant modes; the Wikipedia article below solves the problem for a system of two masses and two springs to obtain two normal modes of oscillation.

6 0

4 years ago