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NemiM [27]
2 years ago
12

The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in

the −x-direction. Vector B⃗ has no x-component.Part A: What is the magnitude of vector B⃗ ? (I solved this question the answer is 3 and it's correct)Part B: What is the direction angle θ of vector B⃗ measured from the +y-direction to the +z-direction? (This is the part that I didn't get it correct)
Physics
1 answer:
g100num [7]2 years ago
8 0

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

  • If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

  • If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

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Where is there kinetic energy in this system?
Alika [10]

Answer:

So kinetic means to move, something like that right, so the two balls that go in the air are where the kinetic energy is.

Explanation:

Hope it helps.

6 0
2 years ago
Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end
iragen [17]

Answer:

The workdone is  W_d =-4400J

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

            The force is on the force gauge  F = 2750 N

             The distance that Magnus pulled the bus  d = 1.60m

Generally  the workdone by the tension force on Magnus is

                  Workdone = Force * displacement \ in \ the \ direction \ of \ force

                     W_d = F * (-d)

This negative sign show that is tension force  is in the opposite direction to Magnus movement (i.e the movement of the bus )

Substituting value we have

                   Workdone  =  - 2750 * 1.60

                                     =-4400 J

7 0
2 years ago
Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
2 years ago
A Styrofoam slab has thickness h and density ρs. When a swimmer of mass m is resting on it, the slab floats in fresh water with
sattari [20]
 <span>A = area of styrofoam 
M = mass of stryofoam = A*h*rho_s 
m = mass of swimmer 

Total mass = m + M = m + A*h*rho_s 
Downward force = g*(total mass) = g*[m + A*h*rho_s] 

The slab is completely submerged. 
Buoyant force = g*(mass of water displaced) = g*[A*h*rho_w] 

Equate these 
g*[m + A*h*rho_s] = g*[A*h*rho_w] 
m + A*h*rho_s = A*h*rho_w 
A*h*[rho_w - rho_s] = m 
A = m/[h*(rho_w - rho_s)]</span>
8 0
2 years ago
A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidentally
vazorg [7]

Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

Explanation:

Using the Bernoulli's laws to use the conserved energy

a. Solve the speed and the radio of this speed of the tank is open to the air

p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 25.68  m²/s²

v1 = 5.06 m/s

Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 15.68  m²/s²

v2 = 3.96 m/s

The ratio on the air is solve using both velocities so:

R = v1/v2 = 5.06 m/s / 3.96 m/s

R = 1.27

b. Now to find the time it takes for the tank to drain if the tank is open to the air

dh/dt = -u

dh/dt = -v * A/A'

dh/dt = v*(.02m)²/(2.0m)² = -v / 10000

and we can further substitute for v:

dh/dt = -(1/1e4)*√[(p+9800h)/500]

Solve replacing

-(1000/49)*√(49000h) = t + C

-(1000/49)*√(49000*0.8) = 0 + C

C = - 4040.6

Then when h = 0,

t = 4286 s

t = 1 hr, 11 min, 26 sec  

6 0
2 years ago
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