If a force of 40N is applied for 0.2 sec to change the momentum of a volleyball, what is the impulse?

A 31 kg crate full of very cute kittens is placed on an incline that is 17° below the horizontal. The crate is connected to a sp

fgiga [73]

Answer:

Explanation:

Change in length of spring = 2.13 m

Component of weight acting on spring = mg sinθ

so

mg sinθ = k x where k is spring constant and x is total stretch due to force on the spring.

Here x = 2.13

mg sin17 = k x 2.13

31 x 9.8 sin17 = k x 2.13

k = 41.7 N/m

b ) In case surface had friction , spring would have stretched by less distance .

It is so because , the work done by gravity in stretching down is stored as potential energy in spring . In case of dissipative force like friction , it also takes up some energy in the form of heat etc so spring stretches less.

5 0

3 years ago

Why pressure above the surface is greater than in the air

nexus9112 [7]

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

7 0

3 years ago

A force of 100 newtons Is applled to a box at an angle of 36° with the horizontal. If the mass of the box Is 25 kilograms, what

faust18 [17]

Horizontal component of force = 100cos(36)= 80.9 N

5 0

3 years ago

¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula

Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

$F = |q|vBsin(\theta)$ (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

$v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s$

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.

Espero que te sea de utilidad!

7 0

3 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire

lana66690 [7]

Answer:

the magnitude of the magnetic force on the wire is 0.2298 N

Explanation:

Given the data in the question;

we know that, the magnitude of magnetic force is given as;

|F $_{mg}^>$ | = I( $B^>$ × $L^>$ )

given that

I = 2.6 A

$B^>$ = 0.17

$L^>$ = 0.52

so we substitute

|F $_{mg}^>$ | = 2.6( 0.17i" × 0.52j" )

|F $_{mg}^>$ | = 0.2298 N

Therefore, the magnitude of the magnetic force on the wire is 0.2298 N

4 0

3 years ago