A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s.
Draw vectors representing the initial velocity final velocity, and (hange in velocity which describe the equation : + Δ¡. 2. The SI nit for weight is the Newton (not the kilogram!). Express the weight of a 2.8 lbs. watermelon in Newtons. 3. On Mars gravitational acceleration is aMars 3.71 m/s2 or aMars 0.37g, where g is the gravitational acceleration of on object on Earth. If a future astronaut drops a rock from a height of 1.9 m how long will it take to hit the ground on Mars? Compare that number to how long the same rock would take to fall on Earth.
2 answers:
Answer:
The time of Mars is 1.65 times larger on Mars than on Earth
Explanation:
The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole
Vf = Vo - V pothole
B) let's transform the weight of free groin system and N international system
1 N = 0.2248 lb
2.8 lbs (1N / 0.2248lbs) = 12.5 N
c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity
Y = Vo t - ½ g t²
Y = - ½ g t²
t = √ 2Y / g
Mars
gm = 0.37g
gm = 0.37 9.8
gm = 3,626 m / s²
t = √( 2 1.9 / 3.626 )
t = 1.02 s
Earth
t = √( 2 1.9 / 9.8)
t = 0.62 s
To make the comparison of time we are the relationship between the two
tm / te = 1.02 / 0.62
tm / te = 1.65
The time of Mars is 1.65 times larger on Mars than on Earth
Answer:
1. The arrows show opposite motion
2 127 N
3. 1 s
Explanation:
1. The arrows will show opposite directions
2. 2.8 lbs = 12.7006 kg
1 kg = 10 N
Therefore, 12.7006 kg = 12.7006 × 10
= 127 N
3. let g = 3.71 m/s²
m = 0.37
h = 1.9
Kinetic energy = mgh
= 0.37 × 3.71 × 1.9
= 2.6 J
The potential energy will be converted to kinetic energy so,
Ek = 1/2 mv
velocity = 3.74 m/s
distance, v
v = u + at
3.74 = 0 + 3.71t
t = 1s
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Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer:
a). 100.2 MHz (typical frequency for FM radio broadcasting)
The wavelength of a frequency of 100.2 Mhz is 2.99m.
b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)
The wavelength of a frequency of 1070 khz is 280.3 m.
c. 835.6 MHz (common frequency used for cell phone communication)
The wavelength of a frequency of 835.6 Mhz is 0.35m.
Explanation:
The wavelength can be determined by the following equation:
(1)
Where c is the speed of light, is the wavelength and
is the frequency.
Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.
<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>
Then, can be isolated from equation 1:
(2)
since the value of c is . It is necessary to express the frequency in units of hertz.
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.
<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>
<em> </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 1070 khz is 280.3 m.
<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.