Answer:
33 Inches
Explanation:
The movement of the ant is sketched in the attached image. The ant moved from the 19-inch mark to 27-inch mark covering a distance of 13 Inches (27 - 19). Finally, it moved from the 27-inch mark to 7-inch mark covering a distance of 20 Inches (27 - 7).
Total Distance covered = 13 + 20 = 33 Inches.
<em>Hence, the total distance the ant traveled is </em><em>33 Inches.</em>
Answer:
The centripetal acceleration at highway speed is greater.
Explanation:
We assume the motion of the car is uniformly accelerated. Let the highway speed be v.
By the equation of motion,
u is the initial velocity, a is acceleration and t is time
Because the car starts from rest, u = 0.
This is the tangential acceleration of the thread of the tire.
The centripetal acceleration is given by
r is the radius of the tire.
Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about
The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.
Answer:
Explanation:
a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.
b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or
c. The period, or T, of a wave is found in the equation
were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:
so
and
T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)
d. The speed of the wave is found in the equation
and since we already have the frequency and we solved for the wavelength already, filling in:
and
v = 50.0(20.0) so
v = 1.00 × 10³ m/s
And there you go!
Answer:
The magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F
Explanation:
Given;
magnitude of the magnetic force is F at 14.0⁰ with respect to the field.
To determine the magnitude of the magnetic force at 32.5⁰ to the field, we apply the following formula and solve with proportion;
f*Sin(14.0⁰) = F
f*Sin(32.5⁰) = ?
Therefore, the magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F