When the ball of the pendulum moves from (x) to (y) in a duration of 0.02 sec the frequency equals

A heating element has a resistance of 180 Ω and draws a current of 1.60 A. What is its power?

iragen [17]

its either 288Watts or 112.5 Watts

5 0

2 years ago

a 0.199 kg snowball moving west makes an inelastic collision with a 2.89 kg box moving 0.523 m/s west. afterward,they move west

kogti [31]

Answer:

The initial velocity of the snowball was 22.21 m/s

Explanation:

Since the collision is inelastic, only momentum is conserved. And since the snowball and the box move together after the collision, they have the same final velocity.

Let $m_1$ be the mass of the ball, and $v_1$ be its initial velocity; let $m_2$ be the mass of the box, and $v_2$ be its velocity; let $v_f$ be the final velocity after the collision, then according to the law of conservation of momentum:

$m_1v_1+m_2v_2=v_f(m_1+m_2)$ .

From this we solve for $v_1$ , the initial velocity of the snowball:

$\boxed{v_1=\frac{v_f(m_1+m_2)-m_2v_2}{m_1}}$

now we plug in the numerical values $m_1=0.199\:kg$ , $m_2=2.89\:kg$ , $v_2=0.523\:m/s$ , and $v_f=1.92\:m/s$ to get:

$v_1=\frac{1.92*(0.199+2.89)-2.89*0.523}{0.199}$

$\boxed{v_1=22.21\:m/s}$

The initial velocity of the snowball is 22.21 m/s.

<em>P.S: we did not take vectors into account because everything is moving in one direction—towards the west.</em>

4 0

2 years ago

Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa

rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

$\theta =1.22\cdot \frac{\lambda }{D}$

Applying the given values we get

$\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads$

thus the linear separation equals $L=\theta \times Distance$

Applying the given values we get

$L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes$

6 0

2 years ago

If a 3.5 gram ping pong ball were traveling to the right horizontally at 12 m/s, and a larger 12 g super ball were thrown direct

algol [13]

Answer:

v = 14.32 m/s

Explanation:

According to the principle of conservation of linear momentum, both the momentum and kinetic energy of the system are conserved. Since the two balls are in the same direction of motion before collision, then;

$m_{1} u_{1}$ + $m_{2} u_{2}$ = ( $m_{1}$ + $m_{2}$ ) v

0.035 × 12 + 0.120 × 15 = (0.035 + 0.120) v

0.420 + 1.800 = (0.155) v

2.22 = 0.155 v

⇒ v = $\frac{2.22}{0.155}$

= 14.323

The velocity of the balls after collision is 14.32 m/s.

3 0

3 years ago

A horse is pulling 53.0 kg plow forward while the ground exerts a 275 N force, and the plow accelerates at 0.222 m/s^2. what is

JulsSmile [24]

Answer:
Magnitude of force the ground exerts on the plow = 263.234 N

Explanation:
Magnitude of force the ground exerts on the plow = Fground - Fplow
We are given that:
Fgound = 275 N
We will now calculate Fplow as follows:
Fplow = mass of horse * acceleration of plow
Fplow = 53 * 0.222
Fplow = 11.766 N

Now, substitute in the above equation to get magnitude of force the ground exerts on the plow as follows:
Magnitude of force the ground exerts on the plow = Fground - Fplow
Magnitude of force the ground exerts on the plow = 275 - 11.766
Magnitude of force the ground exerts on the plow = 263.234 N

Hope this helps :)

3 0

3 years ago