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3 years ago

HELP PLS i need this right now

Physics

1 answer:

N76 [4]3 years ago

7 0

Answer:

V = 44.4 units.

Explanation:

In order to solve this problem, we must use the Pythagorean theorem. Which is defined by the following expression.

$V = \sqrt{(V_{x})^{2} +(V_{y})^{2} }$

where:

Vx = - 43 units

Vy = 11.1 units

Now replacing:

$V=\sqrt{(-43)^{2} +(11.1)^{2} }\\V=44.4 units$

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1. A 1,000-kg car has 50,000 joules of kinetic energy. What is its speed?

bulgar [2K]

Use KE= 1/2mv^2
So...
50,000=(.5)(1,000)v^2
50,000=500 x v^2
Divide 500 on both sides
100 = v^2
Square root both sides to get rid of v^2
Therefore v = 10 m/s

4 0

3 years ago

Read 2 more answers

Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at

sasho [114]

Answer:

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

E_{total} = 2,467 10⁶ N / A θ = -21.8

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

E_total = E₁₃ + E₂₃

bold font vectors . We can work with the components of the electric field in each axis

X- axis

E_ total x = E₁₃ₓ + E_{23x}

y-axis

E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

E = k q / r²

where r is the distance between the charge and the positive test charge

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

E_{total x} = E_{13x} - E_{23x}

E_{total x} = (3.516 - 1.23) 10⁶

E_{total x} = 2.29 10⁶ N / A

E_{total y} = -E_{13y} + E_{23y}

E_{total y} = (-2.777 +1.86) 10⁶ N / A

E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

E_{total} = √ (2.29² + 0.917²) 10⁶

E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

tan θ = E_total and / E_totalx

θ = tan⁻¹ E_{total y} / E_{total x}

θ = tan⁻¹ (-0.917 / 2.29)

θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0

3 years ago

What type of force act on a man at room temperature and in a ground?

lakkis [162]

Isaac Newton’s second law of motion

6 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago

A go kart is traveling at a constant 13.8 meters per second around a horizontal circular track with a radius of 42.0 meters. The

Ivan

Answer:

648.40N

Explanation:

Centripetal acceleration is the acceleration of an object around a circular path. Centripetal for e acting on the combined mass is expressed as:

F = ma

F = mv²/r

m is the mass of the object = 143kg

v is the linear velocity = 13.8m/s

r is the radius = 42.0m

Substitute the given parameters into the formula;

F = 143*13.8²/42

F = 27,232.92/42

F = 648.40N

Hence the magnitude of the centripetal force acting on the combined mass is 648.40N

4 0

3 years ago