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Grace [21]
1 year ago
5

A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the

other side of the peak, as illustrated in Figure 3-29. If the ship shoots the projectile with an initial velocity of v = 248 m/s at an angle of θ = 74°, how close to the enemy ship does the projectile land?
Physics
2 answers:
Ne4ueva [31]1 year ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
valkas [14]1 year ago
3 0

Answer: 2714.24 m

Explanation:

The projectile would have to cross the mountain peak to reach the other side and hit the enemy ship.

Vertical component of velocity = Vy = 248 sin 74° m/s = 238.4 m/s

horizontal component of velocity = Vx = 248 cos 74° m/s = 68.4 m/s

The time taken to reach the peak by the projectile would be half the time taken by projectile to cover the horizontal distance across the mountain.

Using second equation of motion:

Δy = Vy t + 0.5 a t²

⇒0 = 238.4 m/s × t + 0.5 × (-9.8 m/s²) t²

t = 238.4/4.9 = 48.6 s

Horizontal distance covered, x = Vx (t) = 68.4 m/s × 48.6 s = 3324.24 m

It lies at (3324.24 - 610) m = 2714.24 m close to the ship.

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A ball of mass 0.4 kg is initially at rest on the ground. It is kicked and leaves the kicker's foot with a speed of 5.0 m/s in a
yawa3891 [41]

Answer:

the answer the correct  is 3

Explanation:

Let's use the relationship between momentum and momentum

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Let's calculate

         I = 0.4 5.0 - 0

         I = 2.0 N s

By Newton's law of action and reaction the force on the ball is equal to the force that the ball exerts on the foot, therefore the impulse on the foot of equal magnitude, but in the opposite direction

        I = 2.0 Ns with 60°

When reviewing the answer the correct  is 3

4 0
2 years ago
A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

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Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

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so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
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2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer. 

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1 year ago
Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m
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But the value of the mass was previously given, then

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\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}

\frac{A_1}{A_2} = 1

Therefore the ratio of the oscillation amplitudes it is the same.

5 0
2 years ago
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