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2 years ago

7

A body moves in an x-y plane with a velocity vm/s. The component of v are Vx=5m/s, Vy=7m/s along x and y axis respectively. What

is the magnitude and direction of the velocity

1 answer:

mariarad [96]2 years ago

6 0

the answer is in the picture ok

good luck don't worry it's a correct

#carry on learning

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If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is

gavmur [86]

Kelvin is a base unit of temperature scale from SI that defines as zero degree Kelvin (absolute zero). The absolute zero is a hypothetical statement that all molecular movement stops because there is no transient of energy for the molecules to move. When converting temperature in degree Celsius to Kelvin, add 273. You are given 600K and you are asked to find it in degrees Celsius.

T(K) = T(C) + 273
600 K = T(C) + 273
T(C) = 600 – 273
T(C) = 327 °C
<span>The answer is letter B.</span>

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3 years ago

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1. Gas
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4 years ago

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Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

VARVARA [1.3K]

Answer:

Part a)

$f = 4.76 \times 10^{14} Hz$

Part b)

$d = 3.48 \times 10^{-4} m$

Part c)

$\theta = 0.311 degree$

Explanation:

Part a)

As we know that the speed of light is given as

$c = 3 \times 10^8 m/s$

$\lambda = 630 nm$

now the frequency of the light is given as

$f = \frac{c}{\lambda}$

so we have

$f = \frac{3 \times 10^8}{630 \times 10^{-9}}$

$f = 4.76 \times 10^{14} Hz$

Part b)

Position of Nth maximum intensity on the screen is given as

$y_n = \frac{n\lambda L}{d}$

so here we know for 3rd order maximum intensity

$y_3 = 0.76 cm$

n = 3

L = 1.4 m

$0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}$

$d = 3.48 \times 10^{-4} m$

Part c)

angle of third order maximum is given as

$d sin\theta = 3 \lambda$

$3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})$

$\theta = 0.311 degree$

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3 years ago

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Arturiano [62]

Answer:

B. solar flares sunspots, solar flares prominences

Explanation:

Some features of the Sun's surface include sunspots, solar flares, and prominences

pls mark me brainliest its the right answer

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3 years ago

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two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges

Hoochie [10]

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e =8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km$

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