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muminat
3 years ago
12

A proton moves with a speed of 1.17 105 m/s through Earth's magnetic field, which has a value of 50.0 µT at a particular locatio

n. When the proton moves eastward, the magnetic force acting on it is directed straight upward, and when it moves northward, no magnetic force acts on it.
(a) What is the direction of the magnetic field?

(b) What is the strength of the magnetic force when the proton moves eastward? ?N

(c) Calculate the gravitational force on the proton and compare it with the magnetic force. Compare it also with the electric force if there were an electric field with magnitude E = 1.50 102 N/C at that location, a common value at Earth's surface. Note that the mass of the proton is 1.67 10-27 kg. Fgrav = N Felec = N

Suppose an electron is moving due west in the same magnetic field as in the example at a speed of 2.95 105 m/s. Find the magnitude and direction of the magnetic force on the electron. F = ?N
Physics
1 answer:
vlabodo [156]3 years ago
7 0

a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.

b) Given the equation for Magnetic Force

F= qVB

Replacing

F= (1.16*10^{-19})(1.17*10^5)(50*10^{-6})

F=9.36*10^{-19}

c) Given the second Newton's Law by

F_g = 1.67*10^{-27}*9.81

F_g = 1.64*10^{-26}

Given the electric force by,

F_e = 1.6*10^{-19}*1.5*10^2

F_e = 2.4*10^{-17}N

F=9.36*10^{-19}N

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Please help :/ The same motor is used in rockets with different masses. The rockets have different accelerations. According to N
Ymorist [56]

Answer:

1. As rocket mass increases, acceleration decreases.

2. The inverse of the mass of the boat.

Explanation:

1. Newton's second law of motion states;

         F = ma

where F is the force applied, m is the mass and a is the acceleration.

Therefore, increasing the mass of a rocket increases its weight which would reduce its acceleration provided that the force is constant. Thus, as rocket mass increases, acceleration decreases.

2. The slope of the graph can be expressed as;

From Newton's second law,

F = ma

Slope = (Δa) ÷ (ΔF)

Slope = \frac{a}{F}

⇒ \frac{1}{m} = \frac{a}{F}

Therefore, the slope of the graph is the reciprocal of the mass of the boat.

7 0
3 years ago
A 1,200 kg car travels at 20 m/s. what is it’s momentum ?
WINSTONCH [101]

Answer:c  vx v xv xv x

Explanation:

mlmcmcm

6 0
3 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
4 years ago
The tip of the fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotation every
Goryan [66]

Answer:

What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

48 m/s2

53 m/s2

96 m/s2

Answer is 96

Explanation:

5 0
3 years ago
An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00
Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ A = A_0 e^{- \lambda t}

or,

⇒ \frac{A}{A_0}=e^{-\lambda t}

By taking "ln", we get

⇒ ln \frac{A}{A_0}=- \lambda t

By substituting the values, we get

⇒ -ln \frac{110000}{490000} = -48 \lambda

⇒    -1.4939=-48 \lambda

                 \lambda = 0.031122

As,

⇒ \lambda = \frac{ln_2}{\frac{T}{2} }

then,

⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

         =22.27 \ hours  

3 0
3 years ago
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