The Milky Way is the name of the _______ where the Earth is located. A. orbit B. universe C. galaxy D. solar system

A 500g cart moving at 0.25 m/s collides and sticks to a stationary 750g cart. How fast do the two carts

Tresset [83]

Answer:

0.1 m/s

Explanation:

Please see attached photo for explanation.

Mass of 1st cart (m₁) = 500 g

Initial velocity of 1st cart (u₁) = 0.25 m/s

Mass of 2nd cart (m₂) = 750 g

Initial velocity of 2nd cart (u₂) = 0 m/s

Velocity (v) after collision =.?

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(500 × 0.25) + (750 × 0) = v(500 + 750)

125 + 0 = v(1250)

125 = 1250v

Divide both side by 1250

v = 125 / 1250

v = 0.1 m/s

Thus, the two cart will move with a velocity of 0.1 m/s after collision.

3 0

2 years ago

In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the

Levart [38]

Answer:

k = 45.95 N/m

Explanation:

First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

$R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\$

where,

Vo = Launch Speed = ?

R = Horizontal Range = 5.3 m

θ = Launch Angle = 35°

Therefore,

$v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\$

v₀² = 55.33 m²/s²

Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

$Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\$

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

$k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\$

4 0

2 years ago

Wireless Internet networks, including many used in homes, often make use of high-frequency radio waves. High-frequency waves are

lutik1710 [3]

The answer is d , small disruptions

4 0

2 years ago

Read 2 more answers

The linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver

zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force $\times$ Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = $\frac{Impulse}{Time}$ = $\frac{202}{0.244}$ = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

7 0

3 years ago

See the diagram above. Which of the following is the best prediction of what would happen if you increased the distance in secti

Inessa [10]

By the definition of wavelength, the answer is the letter D, the wavelength would decrease.

We can see in the diagram a wave motion.

A wave has some characteristics:

Has an amplitude, the distance from 0 to the crest (highest point in the y-direction, point (3) in the figure) it would see in the figure as (2)
Has wavelength, the distance between the crests.
Has a trough, the lowest point in the y-direction.

Now, if we increase the distance of the crests, by the definition shown above, we will increase the wavelength.

Therefore, the answer is letter D, the wavelength would increase.

You can learn more about wave motion here:

brainly.com/question/22763521

7 0

2 years ago