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3 years ago

A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *

Physics

1 answer:

crimeas [40]3 years ago

8 0

Answer:

Given

mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

p=mv

2kgx3m/s

p=6kg.m/s

the momentum of ball's =6kg.m/s

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Heyi can u answer this matey??

Rudik [331]

It is true my dude.

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3 years ago

Read 2 more answers

¿Qué significa que un automóvil viaje con una rapidez de 45[(Km.)⁄(Hr.)]?

netineya [11]

Responder:

Compruebe amablemente la explicación

Explicación:

¿Qué significa que un automóvil viaje con una rapidez de 45 [(Km.) ⁄ (Hr.)]?

Si se dice que un automóvil viaja a una velocidad de 45 km / h, entonces el automóvil cubre una distancia de 45 km en una hora. 45 km / h es una derivación de la velocidad de desplazamiento obtenida de la relación;

Velocidad = distancia / tiempo

Velocidad = 45 km / 1 h.

Distancia = velocidad × tiempo

Distancia = 45 km / h × 1 h

Distancia = 45 km

Por lo tanto, a esa velocidad, se cubrirá una distancia de 45 km por cada hora que se mueva.

4 0

3 years ago

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

3 years ago

At some instant and location the electric field associated with an electromagnetic wave in vacuum has the strength 96.5 V/m. Fin

prisoha [69]

Answer:

B = 32.17 x 10^-8 Tesla

u = 8.24 x 10^-8 J/m^3

P/A = 24.72 W/m^2

Explanation:

E = 96.5 V/m

velocity of light, c = 3 x 10^8 m/s

Let B be the magnetic field.

The relation between the electric field strength and the magnetic field strength is given by

B = E / c = 96.5 / (3 x 10^8) = 32.17 x 10^-8 Tesla

Let u be the energy density.

$u = \frac{1}{2}\times \varepsilon _{0}E^{2}+\frac{1}{2\mu _{0}}B^{2}$

$u = \frac{1}{2}\times 8.854\times 10^{-12}\times 96.5\times 96.5+\frac{1}{2\times 4\times 3.14\times 10^{-7}}\times 32.17^{2}\times 10^{-16}$

u = 8.24 x 10^-8 J/m^3

Let Power flow per unit area is

P/A = u x c = 8.24 x 10^-8 x 3 x 10^8 = 24.72 W/m^2

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3 years ago

What are the metric units of length?

Anastaziya [24]

Inches, feet, meters

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3 years ago