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Ira Lisetskai [31]
3 years ago
10

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

Physics
1 answer:
VikaD [51]3 years ago
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!
You might be interested in
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:

a)

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.

b)

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

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^
xeze [42]

Answer:

v_{max}=52.38\frac{m}{s}

v_{100}=33.81

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

\sum{F}=0=F_d-W

F_d=W

kv_{max}^2=m*g

v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

\sum{F}=ma=W-F_d

ma=W-F_d

ma=mg-kv_{100}^2

a=g-\frac{kv_{100}^2}{m} (1)

consider the next equation of motion:

a = \frac{(v_{x}-v_0)^2}{2x}

If assuming initial velocity=0:

a = \frac{v_{100}^2}{2x} (2)

joining (1) and (2):

\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}

\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g

v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g

v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}

v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}} (3)

v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}

v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}

v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}

v_{100}=\sqrt{1,143.3}

v_{100}=33.81

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

v = v_0 +at

as stated before, the initial velocity is 0:

v =at (4)

joining (1) and (4) and reducing you will get:

\frac{kt}{m}v^2+v-gt=0

solving for v:

v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }

Plots:

5 0
3 years ago
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