answer.
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]
2 years ago
12

Question #2

Physics
1 answer:
Contact [7]2 years ago
8 0

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

You might be interested in
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magne
Taya2010 [7]

Answer:

Explanation:

a )  No of turns per metre

n = 450 / .35

= 1285.71

Magnetic field inside the solenoid

B = μ₀ n I

Where I is current

B = 4π x 10⁻⁷ x 1285.71 x 1.75

= 28.26 x 10⁻⁴ T

This is the uniform magnetic field inside the solenoid.

b )

Magnetic field around a very long wire at a distance d is given by the expression

B = ( μ₀ /4π ) X 2I / d

= 10⁻⁷ x 2 x ( 1.75 / .01 )

= .35 x 10⁻⁴ T

In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .

6 0
2 years ago
Read 2 more answers
A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene
Ilia_Sergeevich [38]

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

hmax =\frac{vo^{2} }{2*g}

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

5 0
2 years ago
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
2 years ago
Read 2 more answers
The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Elodia [21]

Explanation:

Below is an attachment containing the solution.

7 0
2 years ago
Other questions:
  • A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage
    11·1 answer
  • A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
    9·1 answer
  • Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal
    5·2 answers
  • Energy is observed in two basic forms: potential and kinetic. Which of the following correctly matches these forms with a source
    7·1 answer
  • A box of mass M is pushed a distance Δ x across a level floor by a constant applied force F . The coefficient of kinetic frictio
    12·1 answer
  • Three magnets are placed on a plastic stick as shown in the image. Explain how the magnets need to be rearranged so that they st
    15·2 answers
  • A spring is 14cm long. Three masses are hung from it and then it is measured again. Now it is 19.5cm long. What force did the th
    14·1 answer
  • 8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
    7·1 answer
  • Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
    12·1 answer
  • 3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!