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lozanna [386]
3 years ago
12

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. the heavie

r fragment slides 6.70 m before stopping. how far does the lighter fragment slide?
Physics
1 answer:
Doss [256]3 years ago
6 0
By conservation of momentum,
Distance lighter fragment slide= 7*6.7=46.9m
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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

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2 years ago
The distance between 4 nodes (3 sections, 2 sections= wavelength) is 15.0cm. The frequency of the source is 10Hz. What's the spe
Natalija [7]

Answer:

the speed of the waves is 150 cm/s

Explanation:

Given;

frequency of the wave, f = 10 Hz = 10

distance between 4 nodes, L = 15.0 cm

The wavelength (λ) of the wave is calculated as follows;

Node to Node = λ/2

L = 2(Node to Node) = (4 Nodes) = 2 (λ/2) = λ

Thus, λ = L = 15.0 cm

The speed (v) of the wave is calculated as follows;

v = fλ

v =  10 Hz   x  15.0 cm

v = 150 cm/s

Therefore, the speed of the waves is 150 cm/s

7 0
2 years ago
What is the equation for the enthalpy of vaporization?
Natalka [10]

Answer: Use the formula q = m·ΔHv in which q = heat energy, m = mass, and ΔHv = heat of vaporization.

4 0
2 years ago
A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
Dimas [21]
A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
E=U= \frac{1}{2}kA^2 =  \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J

b) When the position of the object is 
x=1.00 cm = 0.01 m
the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

c) The potential energy when the object is at 
x=3.00 cm=0.03 m
is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
5 0
3 years ago
Tremendous energy is released by nuclear reactions is a measure of the energy required to bond the nucleus together. Where does
Harlamova29_29 [7]

Answer:

Fission is the opposite of fusion and releases energy only when heavy nuclei are split. As noted in Fusion, energy is released if the products of a nuclear reaction have a greater binding energy per nucleon than the parent nuclei.

The amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only 0.1 percent of the original nuclei is converted to energy. Daughter nucleus, energy, and particles such as neutrons are released as a result of the reaction

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