All categories

3 years ago

The magnitude of a vector is indicated by the _____ of the arrow. length direction angle point

Physics

2 answers:

OleMash [197]3 years ago

8 0

Length of the arrow.

tigry1 [53]3 years ago

8 0

the length of a vector arrow represents its magnitude.

You might be interested in

2. An overseas jet requires 6 hours to fly 9700 km. What is the jet's speed?

Debora [2.8K]

Answer:

Most commercial aircraft typically fly at around 460-575 mph, or 740-930 km/h, according to Flight Deck Friend. But private jet speed can vary depending on a variety of factors, such as the weight onboard and the weather conditions.

4 0

2 years ago

What is the acceleration of a car that moves at a steady velocity of 100 km/h for 100 seconds? Explain your answer.

RUDIKE [14]

Answer:

a = 0 m/s²

Explanation:

given,

car moving at steady velocity = 100 Km/h

1 km/h = 0.278 m/s

100 Km/h = 27.8 m/s

time of acceleration = 100 s

acceleration is equal to change in velocity per unit time.

$a=\dfrac{dv}{dt}$

change in velocity of the car is 27.8 - 27.8 = 0

$a=\dfrac{0}{100}$

a = 0 m/s²

If the car is moving with steady velocity then acceleration of the car is zero.

Hence, the acceleration of the car is equal to a = 0 m/s²

3 0

3 years ago

Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth?

aniked [119]

Answer:

Low satellite has high orbital velocity

Explanation:

let v be the orbital speed of the satellite orbiting at a height h is given by

$v=\sqrt{\frac{GM}{R+h}}$

where, M be the mass of planet, r be the radius of planet and h be the height of planet from the surface of planet.

here we observe that more be the height lesser be the orbital velocity.

So, a satellite which is at low height has high orbital velocity.

3 0

3 years ago

An object with a mass of 70 kilograms is supported at a height 8 meters above the ground. What's the potential energy of the obj

alexira [117]

Ep= mgh

70 x 9.8 x 8

Ep= 5,488J

5 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago