All categories

2 years ago

13) Can two nonzero perpendicular vectors be added together so their sum is zero? Explain.

Physics

2 answers:

statuscvo [17]2 years ago

4 0

Answer:

sum of two perpendicular nonzero vectors can not be ZERO

Explanation:

As we know by the sum or addition of two vectors is given by

$R = \sqrt{a^2 + b^2 + 2abcos\theta}$

here we know that

a and b are the magnitude of two vectors

$\theta$ = angle between two vectors

so as we know that here two vectors are perpendicular to each other

so we will have

$R = \sqrt{a^2 + b^2}$

now since the is sum of two positive non zero numbers so it can not be zero

so sum of two perpendicular nonzero vectors can not be ZERO

Neporo4naja [7]2 years ago

3 0

In order for two vectors to add to zero, they must have the same magnitude and point in opposite directions.

Two perpendicular vectors, by definition, make a right angle with each other whereas two vectors pointing in opposite directions form a straight line.

Because of this, two perpendicular vectors with nonzero magnitudes will never add to zero.

You might be interested in

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0

1 year ago

A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Le

Katyanochek1 [597]

Answer:

$\Delta KE = 4.20\times 10^{13}\ J$

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

$\Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})$

$\Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})$

$\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})$

$\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})$

$\Delta KE = 4.20\times 10^{13}\ J$

5 0

2 years ago

The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them

hammer [34]

It is not only the horizontal part of the loop that dips into the magnetic field. We can neglect or disregard the horizontal parts of the loop that hollows into the magnetic fields since only the parts perpendicular to the magnetic field complement to it.

4 0

1 year ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

2 years ago

A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude

Pani-rosa [81]

The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
$y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])$
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
$y(x,t)=0.750 cm$
and therefore this value corresponds to the amplitude of the wave.

4 0

2 years ago

Read 2 more answers