In any electromagnetic wave,

What from the following list of statements about vectors is definitely true? (section 3.3) The magnitude of a vector can be smal

Mashcka [7]

Answer:

"Magnitude of a vector can be zero only if all components of a vector are zero."

Explanation:

"The magnitude of a vector can be smaller than length of one of its components."

Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.

"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."

False. Magnitude of a vector is always positive.

"Magnitude of a vector can be zero if only one of components is zero."

Wrong. For the magnitude of a vector to be zero, all components must be zero.

"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."

Wrong. The magnitude of a vector depends on all components, not only the X component.

"Magnitude of a vector can be zero only if all components of a vector are zero."

True.

6 0

3 years ago

An astronaut who is repairing the outside of her spaceship accidentally pushes away a 92.9 cm long steel rod, which flies off at

DiKsa [7]

Answer:

V = 0.0723 volts = 72.3 milivolts

Explanation:

The emf induced in the rod is the motional emf due to the magnetic field. This motional emf can be calculated by the following formula:

$EMF = V = vBl Sin\theta$

where,

V = Motional EMF = ?

v = speed of rod = 12.5 m/s

B = Magnetic Field = 6.23 mT = 0.00623 T

l = Length of rod = 92.9 cm = 0.929 m

θ = angle between v and B = 90°

Therefore,

$V = (12.5\ m/s)(0.00623\ T)(0.929\ m)Sin\ 90^o\\$

V = 0.0723 volts = 72.3 milivolts

7 0

2 years ago

A

True [87]

Answer:

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

Explanation:

Given:

mass of ice, $m_i=14.7\ g=0.0147\ kg$

mass of water, $m_w=324\ g=0.324\ kg$

Assuming the initial temperature of the ice to be 0° C.

Apply the conservation of energy:

Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

Now from the heat equation:

$Q_i=Q_w$

$m_i.L=m_w.c_w.\Delta T$ ......................(1)

where:

$L=$ latent heat of fusion of ice $=333.55\ J.g^{-1}$

$c_w=$ specific heat of water $=4.186\ J.g^{-1}.^{\circ}C^{-1}$

$\Delta T=$ change in temperature

Putting values in eq. (1):

$14.7 \times 333.55=324\times 4.186\times \Delta T$

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

8 0

3 years ago

A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block

mel-nik [20]

Answer:

v₁ = 0.375 m / s , x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

$p_{f}$ = (m + M) v₁

how the forces are internal

p₀ =p_{f}

m v₀ = (m + M) v₁

v₁ = m /m+M v₀

let's calculate

v₁ = 0.4 /(0.4 + 2.8) 3

v₁ = 0.375 m / s

Let's apply Newton's second law to the Block, to find the friction force

Y axis

N - W = 0

N = W

N = m g

where m is the mass of the block

the friction force has the formula

fr = μ N

fr = μ m g

We apply Newton's second law to slab

X axis

fr = M a

where M is the mass of the slab

μ m g = M a

a = μ g m / M

let's calculate

a = 0.15 9.8 0.4 / 2.8

a = 0.21 m / s²

With kinematics we can find the position

v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

v² = 2 a x

x = v2 / 2a

let's calculate

x = 0.375²/2 0.21

x = 0.335 m

4 0

2 years ago

The materials that made up the solar nebula can be categorized into the four general types as follows. Rank these materials from

san4es73 [151]

Answer:

Hydrogen and Helium
Rock
Metals
Hydrogen compounds

Ranked from left to right based on their abundance in the solar nebula:

Hydrogen and Helium-Hydrogen compounds-Rocks-Metals

I hope you find this information useful and interesting! Good luck!

5 0

3 years ago