All categories

2 years ago

1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package

Physics

1 answer:

Sidana [21]2 years ago

6 0

The velocity of the package after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

and t is time

To calculate the velocity of the package after it has fallen for 3.0 s

That means, we will determine the value of v, at time t = 3.0 s

The parameters are

u = 0 m/s

g = 9.8 m/s²

t = 3.0 s

Putting these values into the equation

v = u + gt

We get

v = 0 + (9.8×3.0)

v = 0 + 29.4

v = 29.4 m/s

Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s

Learn more here: brainly.com/question/13327816

You might be interested in

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

HELP!!!! PLEASE (and there will be more)

Alexus [3.1K]

Answer:

i wanna say tranverse if not surface

Explanation:

6 0

3 years ago

Read 2 more answers

Help with this question please.

mr Goodwill [35]

There are 4 hydrogens on the right side $(2\mathrm H_2=4\mathrm H)$ , and 2 hydrogens on the left per molecule of $\mathrm H_2$ . To get the same number of hydrogens on both sides, the coefficient should be 2.

(Then the number of oxygens will be consistent, since $2\mathrm H_2\mathrm O$ contributes 2 oxygens, and so does $\mathrm O_2$ .)

5 0

3 years ago

What type of rock forms most often near an active volcano?

BaLLatris [955]

Extrusive Igneous Rock

3 0

3 years ago

Read 2 more answers

An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

Ostrovityanka [42]

Answer:

The change in gravitational potential energy of the climber-Earth system is $\Delta PE = 396900 \ J$

Explanation:

From the question we are told that

The mass of the hiker is $m = 75 \ kg$

The time taken is $T = 2 \ hr = 2 * 3600 = 7200 \ s$

The vertical elevation after time T is $H = 540 \ m$

The change in gravitational potential is mathematically represented as

$\Delta PE = mgH$

here g is the acceleration due to gravity with value $g = 9.8 \ m/s^2$

substituting values

$\Delta PE = 75 * 9.8 * 540$

$\Delta PE = 396900 \ J$

3 0

3 years ago