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USPshnik [31]
3 years ago
15

A car is moving eastward and speeding up. the momentum of the car is

Physics
1 answer:
amm18123 years ago
6 0
Accelerating.. I hope that’s the answer your looking for
You might be interested in
Highest common factor of 12r and 10
Scilla [17]

I think the number 2, not sure

3 0
2 years ago
If the displacement of a horizontal mass-spring system was doubled, the elastic potential energy in the system would change by a
bixtya [17]

Answer:

K'=4K

Explanation:

The electric potential energy is given by :

E=\dfrac{1}{2}kx^2

Where

k is spring constant

x is compression or extension in the spring

If the displacement of a horizontal mass-spring system is doubled, x'= 2x

New elastic potential energy :

E'=\dfrac{1}{2}kx'^2\\\\E'=\dfrac{1}{2}k(2x)^2\\\\=\dfrac{1}{2}k\times 4x^2\\\\K'=4\times \dfrac{1}{2}kx^2\\\\K'=4K

So, new elastic potential energy 4 times the initial elastic potential energy.

3 0
2 years ago
A wire is hung between two towers and has a length of 208 m. A current of 154 A exists in the wire, and the potential difference
NARA [144]

Answer:

Mass = 4152kg

Explanation:

Given

L = 208m

I = 154A

V = 0.245V

Density = 3610 kg/m3

ρ = 4.23 x 10-8Ω·m = resistivity of wire

Resistance R = ρL/ A

R = voltage / current = V/I = 0.245/154 = 1.59×10-³ohms

1.59×10-³ = 4.23 x 10-⁸×208/A

Rearranging,

A = 4.23 x 10-⁸×208/1.59×10-³

A = 5.53×10-³m²

Mass = density × volume

Volume = L×A = 208×5.53×10-³m³= 1.15m³

Mass = 3610×1.15 = 4152kg

5 0
2 years ago
Read 2 more answers
Descibe the real-world examples of Newton's third lawthat were idenified in "Applications of Newton's Laws."
ehidna [41]

Answer:

A horse pulls a cart, a person walks on the ground

Explanation:

5 0
2 years ago
Read 2 more answers
1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
3 years ago
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