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adelina 88 [10]
3 years ago
10

Kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth's average speed? (Round your answer to the nearest tenth.) 0.

2 mph 4.8 mph 5.5 mph 144.1 mph
Physics
2 answers:
Natali5045456 [20]3 years ago
6 0

Answer:

4.8mph

Explanation:

Doing it now on EDG

For those doing the exam on EDG science rn, i wish you luck~

Have a good day!~ :D

Anna11 [10]3 years ago
5 0

Answer:

4.8 mph

Explanation:

From the question,

Average speed = total distance/total time

V = d/t....................... Equation 1

Where d = distance, t = time

Given: d = 26.2 miles, t = 5.5 hours.

Substitute these values into equation 1

V = 26.2/5.5

V = 4.76 mph

V ≈ 4.8 mph

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The magnitude of the magnetic force on a current-carrying wire held in a magnetic
Kamila [148]

Answer:

All of the above

Explanation:

The magnitude of the magnetic force on a current-carrying wire held in a magnetic is given by the equation F = BIlsin \theta

Where B = Strength of the magnetic field

I = The current carried by the wire

l = length of the wire in the magnetic field

θ = Angle between the wire and the magnetic field

Based on the relationship written above, the magnitude of the magnetic force on the current - carrying wire in the magnetic field depends on the strength of the magnetic field (B), length of the wire(l), current in the wire (I).

All the options are correct.

3 0
3 years ago
E fundamental frequency of an open organ pipe corresponds to the middle c (261.6 hz on the chromatic musical scale). the third r
luda_lava [24]

The wavelength of the third resonance of the closed organ pipe is equal to the ratio between the speed of sound and the frequency of the 3rd harmonic:

\lambda_3 = \frac{c}{f_3}=\frac{343 m/s}{261.6 Hz} =1.31 m

The relationship between length of a closed pipe and wavelength of the standing waves inside is:

L=\frac{n}{4}\lambda_n

where n is the number of the harmonic. In this case, n=3, so the length of the pipe is

L=\frac{3}{4}(1.31 m)=0.98 m

8 0
3 years ago
The drawing shows a wire composed of three segments, AB, BC, and CD. There is a current of I = 2.0 A in the wire. There is also
alexdok [17]

Answer:

The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.

In fact, the magnetic force exerted by the magnetic field on the wire is

where I is the current in the wire, L the length of the wire, B the magnetic field intensity and  the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is  and so , therefore the magnetic force is zero: F=0.

7 0
3 years ago
A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic
miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by v_{max}=A\omega where A is amplitude and \omega is angular velocity

Angular velocity is given by \omega=\sqrt{\frac{k}{m}}  where k is spring constant and m is mass

So v_{max}=A\sqrt{\frac{k}{m}}

A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m

3 0
3 years ago
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