How much 8 grams will weigh after a physical change

A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed o

Natali5045456 [20]

The time it will take him to fall can be found from:-3m = -(g*t^2)/2
Find that time, it's the time the horse will travel horizontal while the cowboy is falling.So the horizontal distance away is 10 m/s * t

3 0

3 years ago

Evangelista Torricelli was the first person to realize that we live at the bottom of an ocean of air. He correctly surmised that

Radda [10]

Answer:

a) h = 8.02 10³ m b) yes

Explanation:

a) The pressure in a fluid is given by

P = ρ g h

The pressure in this case is the atmospheric pressure, 1.013 105 Pa, let's clear the height (h)

h = P / ρ g

h = 1.013 10⁵ / (1.29 9.8)

h = 8.02 10³ m

b) The height of Mount Everest is 8848 m

It is above this height, according to this model there would be no air to breathe

8 0

2 years ago

Heeeeeeeeeeeeeeeeeelp

RoseWind [281]

Im pretty sure the answer would be thermometer

7 0

2 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

<u>Part A</u>: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$

$v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m.v}{m+m}$

v_1 = v_2 = $\frac{v}{2}$

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = $\frac{2m - m}{2m + m } . v$

v_1 = $\frac{v}{3}$

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

v_1 = v_2 = $\frac{v}{4}$

5 0

3 years ago

A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a

Digiron [165]

The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>

We have the expression for torque acting on a current loop in a uniform magnetic field as,

$\tau=MBsin\theta$

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
Thus, the potential energy will be,

$\tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J$

Thus, we can conclude that, the potential energy will be 1.46*10^-4J.

Learn more about the torque here:

brainly.com/question/27949876

#SPJ4

7 0

1 year ago