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Anastasy [175]
3 years ago
8

The momentum of a 3000 kg truck is 6.36 x 104 kg·m/s. At what speed is the truck traveling? m/s

Physics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

21.2m/s

Explanation

Given data

P=6.36 x 10⁴kg·m/s.

M=3000kg

From the momentum expression

P=mv

v=p/m

Substitute

v=63600/3000

v=21.2m/s

Hence the Velocity is 21.2m/s

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How is air resistance similar to gravity? give me two ways.
OLga [1]

Answer:

1. they both act on an object in free fall

Explanation:

2. both help determine how fast the object will accelerate

4 0
2 years ago
A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn
kow [346]

Answer:

The tension in the cord is T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} } (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

Mg=(M+\frac{MR^{2} }{2b^{2} }  )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2}  }

Replacing in eq. 1

T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

3 0
3 years ago
Read 2 more answers
A 9,100​-microfarad ​[mu​F] capacitor is charged to 22 volts​ [V]. If the capacitor is completely discharged through an iron rod
MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0
3 years ago
28. Sound can be heard around a corner because of
Lesechka [4]

Answer:

Diffraction of sound wavelengths.

Explanation:

Diffraction-A wave is able to bend around a corner due to the effects of diffraction. sound aves are capable of bending around corners in the same magnitude as it's wavelength making it possible to hear sounds around corners.

5 0
3 years ago
An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu
romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

                                   qnet = (Iin - Iout)*t

                                   qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

                                   V = k*qnet / r

Where,                        k = 8.99*10^9   ..... Coulomb's constant

                                   qnet = V*r / k

                                   t = 1000*0.1 / (8.99*10^9 * 0.000002)

                                   t = 5.56 ms

                                   

7 0
3 years ago
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