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3 years ago

14

A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

esistance) the ball will land on the ground with a speed of
A. 0 m/s. B. 20 m/s. C. 30 m/s. D. 40 m/s. E. There is not enough information to say.

2 answers:

ICE Princess25 [194]3 years ago

6 0

B is the answer please mark brainliest

uysha [10]3 years ago

3 0

The answer is not B.

You might be interested in

1. On a force vs. mass graph, what would be the slope of the line?

Luden [163]

1. By Newton's second law,

<em>F</em> = <em>m</em> <em>a</em>

so the slope of the line would represent the mass of the object.

2. If all the forces are balanced, then the object is in equilibrium with zero net force, which in turn means the object is not accelerating. So the object is either motionless or moving at a constant speed.

4 0

3 years ago

A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig

Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

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

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}$

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46

θ₁: is the incident angle = 20.03°

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}$

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.

I hope it helps you!

5 0

3 years ago

Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensiti

Vlad1618 [11]

Answer:

Volume of Sand = 0.4 m³

Radius of Sand Sphere = 0.46 m

Explanation:

First we need to find the volume of gold sphere:

Vg = (4/3)πr³

where,

Vg = Volume of gold sphere = ?

r = radius of gold sphere = 2 cm = 0.02 m

Therefore,

Vg = (4/3)π(0.2 m)³

Vg = 0.0335 m³

Now, we find mass of the gold:

ρg = mg/Vg

where,

ρg = density of gold = 19300 kg/m³

mg = mass of gold = ?

Vg = Volume of gold sphere = 0.0335 m³

Therefore,

mg = (19300 kg/m³)(0.0335 m³)

mg = 646.75 kg

Now, the volume of sand required for equivalent mass of gold, will be given by:

ρs = mg/Vs

where,

ρs = density of sand = 1602 kg/m³

mg = mass of gold = 646.75 kg

Vs = Volume of sand = ?

Therefore,

1602 kg/m³ = 646.75 kg/Vs

Vs = (646.75 kg)/(1602 kg/m³)

<u>Vs = 0.4 m³</u>

Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:

Vs = (4/3)πr³

0.4 m³ = (4/3)πr³

r³ = 3(0.4 m³)/4π

r³ = 0.095 m³

r = ∛(0.095 m³)

<u>r = 0.46 m</u>

4 0

3 years ago

A sodium vapor lamp operates by using electricity to excite the highest-energy electron to the next highest-energy level. light

viva [34]

Answer:

The energy levels are: 3s and 3p

Explanation:

<u>The sodium (Na) element, with atomic number 11 (number of protons: Z=11), has the next electronic configuration</u>:

Z=11: 1s² 2s² 2p⁶ 3s¹ = [Ne] 3s¹

<u>Hence, the electricity used in a sodium vapor lamp will</u> excite the electron of the energy state n=3 from the s-orbital to the next highest energy p-orbital <u>which then</u> drops back to the lower energy s-orbital by the emission of a photon <u>resulting in the production of light</u>.

<u />

<u>The transition of the electron is</u>:

Excitation: [Ne] 3s¹ → [Ne] 3s⁰ 3p¹

Emission: [Ne] 3p¹ → [Ne] 3s¹

So, the energy levels that are involved in the process are 3s and 3p.

Have a nice day!

3 0

3 years ago

A duck has a mass of 2.40 kg. As the duck paddles, a force of 0.150 N acts on it in a direction due east. In addition, the curre

Akimi4 [234]

Answer:

Explanation:

Given

mass of duck=2.40 kg

$F_1$ =0.150 N

$F_2$ =0.160 N in a direction south of east

Net force in x- direction

$F_x=0.150+0.160\cos 45=0.263 N$

$a_x=\frac{0.263}{2.4}=0.109 N$

net force in Y direction

$F_y=0.160\times \sin 45=0.113 N$

$a_y=\frac{0.113}{2.4}=0.047 m/s^2$

Thus displacement in x direction

$x=ut+\frac{at^2}{2}$

$x=0.15\times 2.8+\frac{0.109\times 2.8^2}{2}=0.847 m$

Displacement in Y direction

$Y=ut+\frac{at^2}{2}$

here u=0

$Y=0+\frac{0.047\times 2.8^2}{2}=0.184 m$

net displacement $\sqrt{0.847^2+0.184^2}=\sqrt{0.7513}$

=0.866 m

Direction of displacement

$tan\theta =\frac{0.184}{0.847}$

$\theta =12.25^{\circ}$ south of east

$\theta =-12.25^{\circ}$

3 0

3 years ago