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3 years ago

The four balls have different masses and speeds.

Physics

2 answers:

atroni [7]3 years ago

8 0

<h2>Answer:</h2>

The balls whose velocity and mass is given and momentum of the balls are ranked as C, A, D, B from greatest to least.

<h2>Explanation:</h2>

Momentum is the quantity of physics. Momentum is the product of mass and velocity of the object. Rate of change in momentum is called as force, change in momentum is called impulse. The units of momentum is kg.m/s

Momentum is represented as P which is given by the formula:

$P = m \times a$

Now to calculate the momentums of balls in which speed and mass was given,

Mass = $m_A$ = 0.7 kg, Speed = $a_A$ = 15.0 m/s

Momentum $(P_A)$ = 0.7 $\times$ 15.0

Therefore,

$P_A$ = 10.5 kg.m/s

Momentum of ball A is 10.5 kg.m/s

Mass = $m_B$ = 1.2 kg, Speed = $a_B$ = 5.5 m/s

Momentum $(P_B)$ = 1.2 \times 5.5

Therefore,

$P_B$ = 6.6 kg.m/s

Momentum of ball B is 6.6 kg.m/s

Mass = $m_C$ = 2.5 kg, Speed = $a_C$ = 5.0 m/s.

Momentum $(P_C)$ = 2.5 \times 5

Therefore,

$P_C$ = 12.5 kg.m/s

Momentum of ball C is 12.5 kg.m/s

Mass = $m_D$ = 5.0 kg, Speed = $a_D$ = 1.5 m/s.

Momentum $(P_D)$ = 5.0 $\times$ 1.5

Therefore,

$P_D$ = 7.5 kg.m/s

Momentum of ball D is 7.5 kg.m/s

On comparing the values of momentum,

$C>A>D>B$

g100num [7]3 years ago

5 0

Answer:

C, A, D, B

Explanation:

You just need to calculate the momentum of each ball, by p=m*v

pA = 0.7kg*15m/s = 10.5 kg-m/s

pB = 1.2kg*5.5m/s = 6.6 kg-m/s

pC = 2.5kg*5m/s = 12.5 kg-m/s

pD 5kg*1.5 = 7.5 kg-m/s

Rankin them, we will have: C, A, D, B.

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4 years ago

A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with t

sergij07 [2.7K]

Answer: $135.29\º$

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

<u>*Note the normal is perpendicular to the plane, with a $90\º$ angle with the surface</u>

And this can be visualized in the figure, where the angle ${\theta}_{1}$ with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium refractive index (the air)

$n_{2}=1.309$ is the second medium refractive index (the ice)

$\theta_{1}=25.5\º$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Now, firstly we have to find $\theta_{2}$ and then, by geometry, find $\alpha$ and $\beta$ which sum the <u>angle between the reflected and the refracted light</u>.

Finding $\theta_{2}$ from (1):

$(1)sin(25.5\º)=(1.309)sin(\theta_{2})$

$sin(\theta_{2})=0.328$

$\theta_{2}=arcsin(0.328)$

$\theta_{2}=19.201\º$ (2)

Remembering that the Normal makes a $90\º$ angle with the surface, we can say:

$90\º=\theta_{1}+\beta$ (3)

Finding $\beta$ :

$\beta=90\º-25.5\º=64.5\º$ (4)

Doing the same with $\theta_{2}$ and $\alpha}$ :

$90\º=\theta_{2}+\alpha$ (5)

Finding $\alpha$ :

$\alpha=90\º-19.201\º=70.79\º$ (6)

Adding both angles (4) and (6):

$\alpha+\beta=70.79\º+64.5\º$ (7)

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3 years ago

A shell is shot with an initial velocity v with arrow0 of 18 m/s, at an angle of θ0 = 60° with the horizontal. At the top of the

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Answer:

D = 43 m

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given,

initial velocity = 18 m/s

angle θ = 60°

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$R = \dfrac{v_0^2sin 2\theta}{g}$

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m v₀ cos θ = m₁v₁ + m₂ v₂

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v₂ = 2 v₀ cos θ.

distance covered by the shell from point of explosion

R' = v t

= $(2 v_0 cos \theta) (\dfrac{v_0^2sin \theta}{g})$

= $(2 \dfrac{v_0^2cos \theta sin \theta}{g})$

= $\dfrac{v_0^2sin 2\theta}{g}$

= R

total distance traveled by the shell is

D = $\dfrac{R}{2}+R'$

= 1.5 R

= $1.5\dfrac{v_0^2sin 2\theta}{g}$

D = $1.5\dfrac{18^2sin 2\times 60}{9.81}$

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