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krok68 [10]
3 years ago
5

Salina’s internal desire to exercise is called

Physics
2 answers:
lara31 [8.8K]3 years ago
5 0

the answer your looking for is motivation

zmey [24]3 years ago
4 0
2. Motivation is the internal desire

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Putting a marshmallow directly in the fire
Greeley [361]

Answer:

A conduction is the answer

Explanation:

5 0
2 years ago
A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
dimaraw [331]

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

7 0
2 years ago
An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
A 0.105- kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on
Veseljchak [2.6K]

Answer:

<u><em>0.03 m/s</em></u>

Explanation:

<em>Applying law of conservation of momentum, </em>

  • <em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
  • <em>0.105(24) + 75(0) = (0.105 + 75)v</em>
  • <em>75.105v = 2.52</em>
  • <em>v = 2.52/75.105</em>
  • <em>v = </em><u><em>0.03 m/s</em></u>
3 0
2 years ago
The brakes on a bicycle apply 95 N of force to the wheels. When the
Veronika [31]

Answer:

idk

Explanation:

idk

6 0
3 years ago
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