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Sergio039 [100]

3 years ago

13

A 2 kg box of taffy candy has 40j of potential energy relative to the ground. Its height above the ground is

2 answers:

lord [1]3 years ago

8 0

Answer:

2.04m

Explanation:

PE=mgh

h= PE/mg

h= 40/2(9.82)

h=40/19.64

h=2.04

Alona [7]3 years ago

7 0

Answer:

$20m$

Explanation:

Formula to find the potential energy is,

Potential Energy = mgh

Here,

m = mass

g = gravitational field

h = height

According to the question, we have to find the height.

Usually, we get g is $10ms^{-2}$

Let us solve for the answer now.

$P.E. =mgh\\40J =2kg*10ms^{-2} *h\\40 = 20h\\\frac{40}{20}= \frac{20h}{20} \\20m=h$

Hope this helps you.

Let me know if you have any other questions :-)

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What is the displacement of the runner, whose velocity versus time graph is shown in the Figure, in the first 15.5 s?

muminat

Answer:

10 displacement of the runner

6 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the

Rufina [12.5K]

Answer:

The angular speed is $w = 5.89 \ rad/s$

Explanation:

From the question we are told that

The time taken is $t = 1.6 s$

The number of somersaults is n = 1.5

The total angular displacement during the somersault is mathematically represented as

$\theta = n * 2 * \pi$

substituting values

$\theta = 1.5 * 2 * 3.142$

$\theta = 9.426 \ rad$

The angular speed is mathematically represented as

$w = \frac{\theta }{t}$

substituting values

$w = \frac{9.426}{1.6}$

$w = 5.89 \ rad/s$

3 0

3 years ago

hichkok12 [17]

Gravitational potential energy i think

8 0

3 years ago

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

$s=-8t^2+32t$ ...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

$\dfrac{ds}{dt}=0$

$\dfrac{d(-8t^2+32t)}{dt}=0$

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

$s=-8(2)^2+32(2)$

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0

4 years ago