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Allushta [10]
3 years ago
6

What is the kinetic energy of a 0.135 kg baseball thrown at 40.0 meters per second?

Physics
1 answer:
mixas84 [53]3 years ago
6 0
KE = mass * velocity
^{2}
* 1/2


0.135 * 40.0
^{2}
* 1/2 = 2.7

Answer = 108 J
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A 2.50 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. What is the common velocity of th
yarga [219]

Answer:

V = 2.5*7.0 / ( 2.5 + 70 )

8 0
2 years ago
What is Wave motion<br> las ondas fisica <br> movimiento ondulatorio
PIT_PIT [208]

Answer:

Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties

Explanation:

8 0
2 years ago
What is the mass, in grams, of a pure iron cube that has a volume of 3.70cm cubed?
Arlecino [84]
To determine the amount in grams of the iron, we need data on the density of iron. From literature, it has a value of <span>p=7.9 g/cm3. We simply multiply the volume to the density. We do as follows:

mass = 3.70 (7.9) = 29.23 g Fe

Hope this answers the question. Have a nice day.</span>
7 0
3 years ago
In a car moving at constant acceleration, you travel 230 m between the instants at which the speedometer reads 40 km/h and 70 km
Goryan [66]
The relationship between the distance covered, initial and final speeds, and time can be expressed through the equation,

First equation,

                    2ad = Vf² - Vi²

Substituting the known values,
                   2(a)(0.230 km) = (70 km/h)² - (40 km/h)²
The value of a from the equation is 7173.92 km/h².

Second equation,
                   d = (Vi)(t) + 0.5at²

Substituting the known values,
                    0.230 km = (40 km/h)(t) + (0.5)(7173.92 km/h²)(t²)

The value of t from the equation is 4.1818 x 10^-3 hours which is also equal to 0.2509 minutes or 15 seconds.

Answer: 15 seconds
7 0
3 years ago
The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e
Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed (v), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi (1)

Where:

\Delta t - Rotation time, measured in seconds.

R - Radius of the Earth, measured in meters.

\phi - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that \Delta t = 86160\,s, R = 6.371\times 10^{6}\,m and \phi = 37.6819^{\circ}, then the tangential speed at Livermore is:

v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}

v\approx 284.001\,\frac{m}{s}

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0
3 years ago
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