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navik [9.2K]
3 years ago
14

A 700 kg racecar slowed from 30 m/s to 15 m/s. What is the change in momentum

Physics
1 answer:
Lerok [7]3 years ago
8 0

Momentum = (mass) x (speed)

Original momentum  =  (700 kg) x (30 m/s)  =  21,000 kg-m/s

Final momentum = (700 kg) x (15 m/s)  =  10,500 kg-m/s

Change in momentum  =  - 10,500 kg-m/s .
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Two long, straight wires are parallel and 26 cm apart.
mezya [45]

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

= 4080.744×10^-6/ 1.634

= 4.080×10^-3 / 1.634

= 2.49×10^-3 N/m

5 0
3 years ago
What is the definition of the word amplitude
qwelly [4]

Explanation:

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path. ... Waves are generated by vibrating sources, their amplitude being proportional to the amplitude of the source.

8 0
2 years ago
Which is the smallest atom? A. Magnesium (Mg) B. Calcium (Ca) C. Barium (Ba) D. Strontium (Sr)
padilas [110]

Try option B or option A

6 0
3 years ago
Walt ran 5 km in 25 minutes going east to what was his average velocity
goldenfox [79]

distance d = 5 km = 5 x 1000 m = 5000 m

time taken = 25 minute = 25 x 60 sec = 1500 sec

average velocity V = d/t

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3 0
3 years ago
Read 2 more answers
A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
Korvikt [17]

Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}

This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

8 0
3 years ago
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