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astra-53 [7]
2 years ago
9

A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

is needed to select a speed of 4.40 ✕ 106 m/s? V/m
Physics
1 answer:
Alekssandra [29.7K]2 years ago
4 0

Answer:

The electric field strength is 6.6\times10^{5}\ V/m

Explanation:

Given that,

Magnetic field = 0.150 T

Speed v= 4.40\times10^{6}\ m/s

We need to calculate the electric field strength

Using formula of velocity

v=\dfrac{E}{B}

E=v\times B

Where, v = speed

B = magnetic field

Put the value into the formula

E=4.40\times10^{6}\times0.150

E=660000\ V/m

E=6.6\times10^{5}\ V/m

Hence, The electric field strength is 6.6\times10^{5}\ V/m

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Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jump up again to celebrat
Ludmilka [50]

Answer:

1800 N

Solution:

Impulse = mΔv = m * (u - v) .

 here m = 100 kg

          u = 4 m/s

          v = -5 m/s

  impulse = 100 x ( 4 - ( -5 ) ) = 900 Kg m/s .

Average reaction Force ( Favg ) = impulse / Δt

Average reaction Force ( Favg )  = 900kg·m/s / 0.5s

Average reaction Force ( Favg )  = 1800 N

6 0
2 years ago
What are the methods of heat transfer? ​
m_a_m_a [10]

Answer:

Radiation , Conduction and Convection

Explanation:

Those are the ways heat is transferred

8 0
3 years ago
A 50 kg bumper car with a 40 kg child and it is at rest when a 60 kg child in her own bumper car slams into it the collision las
IrinaK [193]

Answer:

 F = 99 v₂₀

v₂₀ = 1 m / s,        F = 99 N

Explanation:

In this exercise it is asked to find the force during the collision, for this we use the relationship between the momentum and the momentum of car 1

            I = Δp

            F t = p_f- p₀

            F t = m (v_f -v₀)                        (1)

We must find the final speed of car 1, for this we define a system formed by the two cars, in this case the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = 0 + m₂ v₂₀

         

final instant. After the crash

        p_f = m₁ v₁ + m₂ v_{2f}

the moment is preserved

        p₀ = p_f

        m₂ v₂₀ = m₁ v_{1f} + m₂ v_{2f}           (2)

        m₂ (v₂₀ - v_2f}) = m₁ v_{1f}

as the collision is elastic the kinetic energy is also conserved

        K₀ = K_f

        ½ m₂ v₂₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

        m₂ (v₂₀² -v_{2f}²) = m₁ v_{1f}²

let's write our system of equations, using

         a² - b² = (a + b) (a-b)

         m₂ (v₂₀ - v_{2f}) = m₁ v_{1f}

         m₂ (v₂₀ -v_{2f}) (v₂₀ + v_{2f}) = m₁ v_{1f}²

to solve we divide the equations

       v₂₀ + v_{2f} = v_{1f}

with this we substitute in equation 2 and find the speed of each car, in this case we need the speed of car 1

         m₂ v₂₀ = m₁ v_{1f} + m₂ (v_{1f}-v₂₀)

         2m₂ v₂₀ = (m₁ + m₂) v_{1f}

          v_{1f} = \frac{2m_2}{m_1+m_2}  v_{2o}

We substitute in the drive ratio of car 1

            F t = m (v_f -v₀)

            F = m₁ (\frac{2m_2}{m_1+m_2}  v_{2o} - 0) / t

            F = \frac{2m_1 m_2 }{m_1+m_2}   \   \frac{v_{2o}}{t}

the mass of each car is the mass of the car plus the mass of the boy

           m₁ = 50 +40 = 90 kg

           m₂ = 50 +60 = 110 kg

     

time is t = 1

         

we substitute the values

           F = \frac{ 2\  90 \ 110}{90+110}  \ \frac{v_{2o}}{1}2 90 100/90 + 110 vo2 / 1

           F = 99 v₂₀

The value of the initial velocity of car 2 is not indicated in the problem, if this velocity is known it can be included and the force value is obtained, suppose that the initial velocity v₂₀ = 1 m / s

           F = 99 N

4 0
3 years ago
Time Warner Cable's leadership development program that spanned over 30 days and included weekly videos, practice exercises, and
ankoles [38]

Answer: A.

tracking training through a leaming records store LRS.

Explanation:

An LRS uses xAPI to collect learner data, or experiences, from both online and offline sources. These experiences are reported back to the LRS in the form of xAPI statements, where they are stored. These statements can then be retrieved for reporting and interpretation of the learner data.

8 0
2 years ago
A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0
2 years ago
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