A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

Answer 1

Answer:

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) $h*(1 - 1/2 g * h/v_0^2)$

b) $h = v_0^2/ g$

c) $h = v_0^2/ g$

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

$y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t$

where:

y = height at time t

y0 = initial height

v0 = initial velocity

a = acceleration

t = time

v = velocity

a) When the balls collide, h1 = h2. Then,

$h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0$

Replacing in the equation of the height of the first ball:

$h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)$

b) that the balls collide at t = h/v0. Then:

$h/ v_0 = -v_0/-g\nh = v_0^2/ g$

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

$h = v_0^2/ g$

See more about velocity at brainly.com/question/862972

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What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

The mass of a neutron is:
$m=1.67 \cdot 10^(-27)kg$
Since we know its speed, we can calculate the neutron's momentum:
$p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s$

The problem says the photon has the same momentum of the neutron, p. The photon momentum is given by
$p= (h)/(\lambda)$
where h is the Planck constant, and $\lambda$ is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
$\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m$

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
$f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz$

With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m

Answers

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² = 2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

How much electrical energy is used by a 400 W toaster that is operating for 5minutes?
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J

Answers

The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

What is the power output?

The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.

The given data in the problem is;

E is the electrical energy

P is the power output = 400 W

t is the time period = 5 minutes

The power output is given as;

$\rm P= (E)/(t) \n\n\ E= P * t \n\n\ E= 400 * 300 \n\n\ E=120,000 \ J$

Hence the electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

To learn more about the power output refer to the link;

brainly.com/question/22285866

Answer:

The answer is C. 120,000 J.

Explanation:

You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You place the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces

Answers

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that $v = f * d$ where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.