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3 years ago

How does the mass affect the speed as it rolls down a slope?

1 answer:

6 0

As we use the Kinetic energy and the equation is 1/2mv^2, changing its mass will change its speed and its energy. So more mass, more speed more energy. also the gravitational potential energy; mass x gravity x height; more mass and more height more speed as it go down to the slope! Hope it helps!

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Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1

My name is Ann [436]

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

$\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons} = 4.41294 Mev/nucleon$

So approximately 4.41294 Mev/necleon

3 0

2 years ago

John and Caroline go out for a walk one day. This graph represents their distance from home.

Serga [27]

Answer:

Option C is the correct answer.

Explanation:

We have velocity of a body = Change in position/ Time.

Considering first portion of graph,

Change in position = 30 - 0 = 30 m

Time = 0.75 hours = 45 minutes = 2700 seconds

Velocity = 30/2700 = 0.011 m/s

Considering second portion of graph,

Change in position = 30 - 30 = 0 m

Time = 0.5 hours = 30 minutes = 1800 seconds

Velocity = 0/1800 = 0 m/s

Considering third portion of graph,

Change in position = 0 - 30 = -30 m

Time = 0.75 hours = 45 minutes = 2700 seconds

Velocity = -30/2700 = -0.011 m/s

So firstly they walked in one direction(positive direction), then they were still(velocity is zero), then they walked in the opposite direction( velocity is negative).

Option C is the correct answer.

3 0

3 years ago

Read 2 more answers

Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and

Liono4ka [1.6K]

a) $5 \Omega$ , 1.6 A

b) $6 \Omega$ , 1.33 A

Explanation:

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

$R=R_1+R_2$

where

$R_1=4\Omega$

$R_2=1 \Omega$

Therefore, the equivalent resistance is

$R=4+1=5 \Omega$

Now we can use Ohm's Law to find the current flowing through the circuit:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=5\Omega$ is the equivalent resistance of the circuit

Substituting,

$I=\frac{8}{5}=1.6 A$

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

$R=R_1+R_2+R_3$

where:

$R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega$

Substituting, we find the equivalent resistance:

$R=4+1+1=6 \Omega$

Now we can find the current through the circuit by using again Ohm's Law:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=6\Omega$ is the equivalent resistance

Substituting,

$I=\frac{8}{6}=1.33 A$

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0

3 years ago

A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p

Anettt [7]

Answer:

A u = 0.36c B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

u’ (1- u v / c²) = u -v

u + u ’uv / c² = v - u’

u (1 + u ’v / c²) = v - u’

u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

u = 0.51c / (1 + 0.4042)

u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c