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Wittaler [7]
3 years ago
12

A bird flies from the South Pole to the North Pole.part of the journey is 1,000 miles that takes 2 weeks,what is the birds veloc

ity in that time,what is the answer
Physics
1 answer:
avanturin [10]3 years ago
7 0
1000 miles = 1 609 340 m
2 weeks = 1209 600 s
v = 1609340/1209600 = 1.33 m/s
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A 30.4-newton force is used to slide a 40.0-newton crate a distance of 6.00 meters at constant speed along an incline to a verti
Angelina_Jolie [31]

Hi there!

Since the crate is being slid at a constant speed, the forces sum to 0 N. In this instance, the following forces occur in the axis of interest:

Wsinθ = downward acceleration along incline due to gravity (N)

Fκ = kinetic friction force along incline (N)

A = applied force (N)

The acceleration due to gravity and friction force act in the same direction, so:

Wsinθ + Fκ = A

Solve for sinθ using right triangle trigonometry:

sinθ = O/H = 3/6 = 0.5

Rearrange the equation for the force of kinetic friction and solve:

Fκ = A - 0.5W

Fκ = 30.4 - 20 = 10.4 N

Now, recall that:

Work = Force × displacement (W = F × d)

Since the box's displacement is in the same axis as the force but OPPOSITE direction, we must use:

W = Fdcosθ

Angle between displacement and friction force is 180°.

cos(180) = -1

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3 years ago
Which step of the scientific methood do you perform after you state the problem
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After stating the problem, we form a hypothesis, or an explanation that can be tested.

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An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi
Sergio039 [100]

Answer:

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(b) 5.056 x 10^-14 N

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Y component of velocity of electron is 2.4 × 10^6 m/s

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Y component of magnetic field is  -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}

\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}

The force on the charge particle when it is moving in the magnetic field is given by

\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )

(a) Force on electron is given by

\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )

\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}

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(b) Force on a proton is given by

\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )

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Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

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