1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
spin [16.1K]
4 years ago
7

It takes a student 12.0 minutes to get home from school while riding their bike at 6.7 m/s.

Physics
1 answer:
givi [52]4 years ago
4 0

Answer: 289.44km

Explanation:

You might be interested in
A sledgehammer hits a wall How do the hammer and the wall act on each other?
tigry1 [53]

We want to study the impact of a sledgehammer and a wall.

Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.

When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.

This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.

Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.

If you want to learn more, you can read:

brainly.com/question/13952508

7 0
3 years ago
A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0
3 years ago
1. Una bola de 0.510 kg de masa se mueve al este (dirección +x) con una rapidez de 4.80 m/s y choca frontalmente con una bola de
Grace [21]

Responder:

3,37 m / s, + ve x - dirección

Explicación:

Utilizando la ley de conservación de la cantidad de movimiento expresada por la fórmula;

m1u1 + m2u2 = (m1 + m2) v

m1 y m2 son las masas de los objetos

u1 y u2 son sus velocidades iniciales

v es su velocidad común

Dado

m1 = 0,519 kg

u1 = 4,80 m / s

m2 = 0,220 kg

u2 = 0 m / s (cuerpo en reposo)

Necesario

Velocidad común v

Sustituir en la fórmula los valores dados;

0,519 (4,8) + 0,22 (0) = (0,519 + 0,220) v

2,4912 + 0 = 0,739 v

2,4912 = 0,739v

Dividir ambos lados por 0,739

2,4912 / 0,739 = 0,739 / 0,739

<em>v = 3,37 m / s </em>

<em>Por lo tanto, la rapidez de ambas bolas después de la colisión es de 3.37 m.s hacia la dirección x positiva, ya que m1> m2 y la velocidad común es positiva.</em>

6 0
3 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
A(n ___ is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions.
Fynjy0 [20]
A stereotaxic atlas is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions. A stereotaxic atlas is used for doing a stereotactic surgery in order to cure the disorder occurred in the brain. This tool increase the accuracy, reliability, and the measurement of the brain because it gives the accurate brain map<span>.</span>
5 0
3 years ago
Other questions:
  • Which element has the greatest number of valence electrons?
    12·2 answers
  • Our Sun is considered a(n) _____ star.
    12·1 answer
  • Which involves reflection of light waves? A. A full moon shows sunlight bouncing off the moon's surface and traveling toward Ear
    8·2 answers
  • Rounded to the nearest whole number, how many electrons are in an atom of zirconium? 40 51 91 131
    8·2 answers
  • Friction exists only when two objects rub against each other.<br> A.True<br> B.False
    14·2 answers
  • Please help, which vector represent the force that will produce equilibrium with these two forces?
    13·1 answer
  • if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use
    8·1 answer
  • Based on the solubility graph. what effect does increasing the temperature of these substances have in their solubility? does th
    7·1 answer
  • A camel is living in an oasis(origin), but wants to travel to a new oasis. To do this the camel travels 3,000 meters east, and t
    14·1 answer
  • A bullet is fired horizontally at 343 m/s from the top of a building where height is 37.3 m. The built will be
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!