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3 years ago

What are the components of the vector with a magnitude of 65.0 and a direction of 101.7°

Physics

2 answers:

artcher [175]3 years ago

8 0

The correct answer is (63.6, -13.2)

soldier1979 [14.2K]3 years ago

3 0

Answer:

The components of the vector = (-13.18i, 63.65j)

Explanation:

Given;

magnitude of magnetic field strength, B = 65.0

direction of the magnetic field strength, θ = 101.7°

The vector has both y-component and x-component

The y-component of the vector is given as, Bsinθ,

The x-component of the vector is given as, Bcosθ,

The components of the vector = (x, y) = ( Bcosθ, Bsinθ)

The components of the vector = ( 65cos101.7°, 65sin101.7°)

The components of the vector = (-13.18i, 63.65j)

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Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What

yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0

3 years ago

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

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Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

$F_n=F-F'$

$F_n=37.2-22.4$

$F_n=14.8\ N$

Work done on the wagon is given by the product of net force and displacement. It is given by :

$W=F_n\times d$

$W=14.8\ N\times 2.5\ m$

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

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pav-90 [236]

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4 years ago

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