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3 years ago

What kind of energy does a rubber band have when it is stretched?

Physics

2 answers:

dalvyx [7]3 years ago

6 0

Answer:

elastic potential energy

Explanation:

RideAnS [48]3 years ago

5 0

Elastic potential energy. When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.

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How do shows and movies affect our lives? cons. answer honestly.:)

bezimeni [28]

We getting addicted so it keeps us from getting bored

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3 years ago

Some cities now implement signal lights designed to specifically apply to _____ rather than motorized vehicles.

dmitriy555 [2]

The correct answer is C; Bicycles.

Further Explanation:

In major cities, in the United States, have implemented signal lights specifically designed for bicycle riders. The riders also have their own designated bike lanes in many large cities. Drivers in vehicles, are to give the right of way to people on bicycles.

Bicycle riders are to follow the same laws and laws specifically for the riders or they can face fines and tickets.

Learn more about bicycle laws at brainly.com/question/8934107

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2 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

2 years ago

2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (

DiKsa [7]

Answer with Explanation:

We are given that

$D_1=0.15mGy$

$D_2=0.22mGy$

$Q_1=6$

$Q_2=10$

a.We have to find the total dose

Total dose= $D=D_1+D_2$

Using the formula then, we get

$D=0.15+0.22$

$D=0.37mGy$

b.We have to find the total dose equivalent

Total dose equivalent=H= $D_1Q_1+D_2Q_2$

Using the formula

$H=0.15(6)+0.22(10)$

H=3.1mSv

7 0

2 years ago

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$