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3 years ago

6

please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ba

ll changed to -0.4 meter/second. The collision takes 0.2 seconds to occur. What’s the acceleration of the ball during the collision? Use . a= v-u/t

1 answer:

viva [34]3 years ago

4 0

Answer:

the acceleration during the collision is: - 5 $\frac{m}{s^2}$

Explanation:

Using the formula:

$a=\frac{\Delta\,v}{\Delta\,t}$

we get:

$a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}$

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PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball

bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

q₁ = 3C

q₂ = 14C

Distance between balls = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

F = $\frac{k q_{1} q_{2} }{r^{2} }$

where k = 9 x 10⁹Nm²/C²

q is the charges

r is the distance

Input the variables and solve;

F = $\frac{9 x 10^{9} x 3 x 14 }{9000}$ = 4.2 x 10⁷N

8 0

3 years ago

Use the following photoelectric graph to answer the following question:

klio [65]

The photoelectric effect is obtained when you shine a light on a material, resulting in the emission of electrons.

The kinetic energy of the electrons depends on the frequency of the light:
K = h(f - f₀)

where:
K = kinetic energy
h = Planck constant
f = light frequency
f₀ = threshold frequency

Rearranging the formula in the form y = m·x + b, we get:
K = hf - hf₀
where:
K = dependent variable
f = <span>indipendent variable
h = slope
hf</span>₀ = y-intercept

Every material has its own threshold frequency, therefore, what stays constant for all the materials is h = Planck constant (see picture attached).

Hence, the correct answer is C) the slope.

5 0

3 years ago

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 24 m/

PilotLPTM [1.2K]

Answer:

a) Vf = 27.13 m/s

b) It would have been the same

Explanation:

On the y-axis:

$Y=-Vo*sin\theta*t-1/2*g*t^2$

$-8=-24*sin(21)*t-1/2*10*t^2$

Solving for t:

t1 = 0.67s t2= -2.4s

Discarding the negative value and using the positive one to calculate the velocity:

$Vf_y = -Vo*sin\theta-g*t$

$Vf_y = -15.3m/s$

So, the module of the velocity will be:

$Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}$

$Vf=27.13m/s$

If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.

3 0

3 years ago

which one of the following is not an example of a combination reaction a. 2H2O(I)---> 2H2(g)+O2(g) b. 2K+CI2--->2KCI c. 2H

Alexxandr [17]

The answer would be A.

Combination reaction, otherwise known as synthesis reaction, occurs when two or more substances react and form a single, more complex substance.

A combination reaction looks like this in an equation:

<em> A + B </em><em>→ </em><em>AB</em>
reactants product

So from two reactants they become one product.
Let's take a look at choice A.

2H₂O(l) → 2H₂(g) + O₂(g)
reactant product(s)

Notice that from one substance, it broke down into two substances. So this is not a combination reaction. This is what you call a <em>decomposition reaction. </em>

6 0

2 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

Lyrx [107]

Answer:

$\theta=4.64^{\circ}$

Explanation:

It is given that,

The frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Slit separation, $d=2.2\times 10^{-5}\ m$

Let $\theta$ is the angle away from the central bright spot the third bright fringe past the central bright spot occur. The condition for bright fringe is :

$d\ sin\theta=n\lambda$

n = 3

$\lambda=\dfrac{c}{f}$

$d\ sin\theta=\dfrac{nc}{f}$

$sin\theta=\dfrac{nc}{fd}$

$sin\theta=\dfrac{3\times 3\times 10^8}{5\times 10^{14}\times 2.2\times 10^{-5}}$

$\theta=4.64^{\circ}$

So, at 4.64 degrees the third bright fringe past the central bright spot occur. Hence, this is the required solution.

3 0

2 years ago