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Vikki [24]
4 years ago
8

A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the

rubber band?
Physics
2 answers:
Valentin [98]4 years ago
7 0
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m

Therefore, the displacement would be 0.32 m.
Allisa [31]4 years ago
7 0
The correct answer is 0.4 m on Edge.
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MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo
liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}

hence, the mean velocity is 8.4 m/s

8 0
3 years ago
What two measurements are needed to determine density?
g100num [7]
Density is defined as (mass) per unit (volume).  So in order to calculate
the density of a glob of some substance, you pretty much have to measure
its mass and its volume. 
7 0
3 years ago
Read 2 more answers
Which statement best describes a characteristic of gases?
salantis [7]
Assumes the shape and volume of its container 
<span>particles can move past one another</span>
8 0
3 years ago
Find the weight of a 25 kg table. (Use g= 10 m/s?)
PtichkaEL [24]

Answer:

weight = 25*10 =250 N

Explanation:

g must be given in units of m/s^2

The weight of any type of body will be the product of his mass by the gravity

where:

m =mass [kg]

F = force [N] or [kg*m/s^2]

g = acceleration [m/s^2]

3 0
4 years ago
Read 2 more answers
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
3 years ago
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