Halogens (Group 17) have<br> valence electrons and will<br> to become stable

A wave is traveling at a speed of 18 m/s with a frequency of 3 Hz. A second wave is traveling at a speed of 16 m/s with a freque

Alik [6]

Answer:

2Hz

Explanation:

8 0

3 years ago

A 1.5kg cannon is loaded with a 0.52 kg ball . The cannon is ignited and launches the ball forward with speed of 75m/s. Determin

Mila [183]

Answer:

-26 m/s (backward)

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, the total momentum momentum of the cannon + ball system must be conserved before and after the explosion.

Before the explosion, they are both at rest, so the total momentum is zero:

p = 0

After the explosion, the total momentum is:

$p=MV+mv$

where

M = 1.5 kg is the mass of the cannon

m = 0.52 kg is the mass of the ball

v = +75 m/s is the velocity of the ball

V is the velocity of the cannon

Since the momentum is conserved, we can equate the two expressions:

$0=MV+mv$

And solving, we find V:

$V=-\frac{mv}{M}=-\frac{(0.52)(+75)}{1.5}=-26 m/s$

where the negative sign means the direction is opposite to that of the ball.

4 0

4 years ago

John rides his bike with a constant speed of 12 miles per hour. How far can he travel in 2 hours?

Gwar [14]

Answer:

24 miles

Explanation:

Make a proportion

He is traveling at a constant rate of 12 miles/hour, so

he can go 12 miles in 1 hour, and x miles in 2 hours

12/1=x/2

Multiply both sides by 2

2*12/1=x/2 *2

24=x

So, in 2 hours he can travel 24 miles

5 0

3 years ago

The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

Mamont248 [21]

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

$a_c = \frac{v^2}{R}$

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

$v = R\omega$

Where,

R = Radius

$\omega =$ Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

$\omega = \frac{2\pi}{T}$

T = Period

So replacing we have to

$\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s$

Since $1 Light year = 9.48*10^{15}m$

Then the radius in meters would be

$R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})$

$R = 2.844*10^{20}m$

Then the centripetal acceleration would be

$a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2$

From the result obtained, considering that it is an unimaginably low value of an order of less than $10^{-10}$ it is possible to conclude that it supports the assertion on the inertial reference frame.

8 0

3 years ago

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floo

Nuetrik [128]

Answer:

a) v = 0

b) The aceleration is 1.41 $m/s^{2}$

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

$\sum{F} = 0 = F_{spring} - F_{external}$

$F_{spring} = F_{external}$

$Kx = F_{external}$

$x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm$

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

$\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma$

For the y axis:

$F_N = W = mg = 3*9.8 = 29.4N$

To calculate the force of friction:

$F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N$

Therefore for x axis:

${Kx-F_{external}-F_{friction}}=ma$

$a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}$

7 0

3 years ago

Halogens (Group 17) have valence electrons and will to become stable