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LenaWriter [7]
3 years ago
5

If C moves to position 5 (60 meters), what is its average velocity during these 5 seconds?

Physics
2 answers:
Zina [86]3 years ago
4 0

The average velocity is simply the change in displacement over the specified time interval. In this case we have a total of five intervals, which represents 5 seconds. Dividing 60 m by 5s, gives us 12 m/s, which is the magnitude of the average speed in the interval.

Hoochie [10]3 years ago
3 0

Velocity:

60 meters/ 5 seconds= 12 meters/second

Velocity=12meters/second

(60 meters divided by 5 seconds equals 12 meters per second)

Hope this helps!

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A bullet with a mass of 4.26 g and a speed of 881 m/s penetrates a tree horizontally to a depth of 4.44 cm. Assume that a consta
sveta [45]

Let F be the magnitude of the frictional force. This force performs an amount of work W on the bullet such that

W = -Fx

where x is the distance over which F is acting. This is the only force acting on the bullet as it penetrates the tree. The work-energy theorem says the total work performed on a body is equal to the change in that body's kinetic energy, so we have

W = ∆K

-Fx = 0 - 1/2 mv²

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Solve for F and plug in the given information:

F = mv²/(2x)

F = (0.00426 kg) (881 m/s)² / (2 (0.0444 m))

F = 37,234.8 N ≈ 37.2 kN

8 0
2 years ago
ILL MARK BRAINLIEST
77julia77 [94]

Answer: B

Longitudinal wave

Explanation:

Transverse waves have crests and troughs

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Since sound wave is a longitudinal wave. And longitudinal wave exists apart from sound, we can therefore conclude that it's a longitudinal wave in spring.

8 0
3 years ago
If the energy of a photon is 1.32 × 10¯18 j, what is its wavelength in nm?
Katarina [22]
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3 years ago
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A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release
iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= 2pie\frac sqrt {m}{k}

 \frac {{f2}/times {fo}}=1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒\frac{f2} {f1} = \frac sqrt{m1}[m2}

⇒\frac{f2}{fo} = 1:3

⇒f2=\frac{fo} {3}

6 0
3 years ago
2. A 50 kg diver is standing on the edge of a 15 m high cliff. What is his potential energy?
Lera25 [3.4K]

Answer:

3675 J

Explanation:

Gravitational Potential Energy = \frac{1}{2} × mass × g × height

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Mass = 50 kg

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Height = 15 m

Gravitational Potential Energy = \frac{1}{2} × 50 ×9.8 × 15

= 3675 J

3 0
3 years ago
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