Ask question

Ask question

All categories

2 years ago

12

A 65 kg bungee jumper jumps off of a bridge. At an instant when they are 50 m above the ground, they are falling at a speed of 4

0 m/s, and the elastic in their bungee cord
(k = 650 N / m) is stretched by a distance of 2.0 m. What is the total mechanical energy of the bungee jumper at this instant ?
a. 1300 J
b. 52000 J
c. 31850 J
d. 85150 J

2 answers:

enot [183]2 years ago

8 0

The answer is 1 because u just did this test

maw [93]2 years ago

3 0

Answer:

the mechanical energy = a 1300J

You might be interested in

When a wave passes from one medium to another, its _________ remains constant.

Mashcka [7]

The frequency will not change

3 0

3 years ago

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m= $1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}$ = $7*10^{14}$ m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

$X=u[2y/a]^{1/2}$

= $4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}$

= $3*10^{-2}$ = 3 cm

5 0

4 years ago

One disadvantage of using proprietary licensed software is that

Vaselesa [24]

Answer:

its b

Explanation:

5 0

3 years ago

Read 2 more answers

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago