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2 years ago

SORRY LAST ONE FOR TODAY, IM GOING TO PASS IF I GET THIS ANSWERED

Physics

2 answers:

faust18 [17]2 years ago

6 0

Answer:

According to the Given Statement 

Canon fires directly upwarda at 150m/s ( Initial speed, u )

Acceleration due to gravity is 10m/s²

Since Cannon ball reach it maximum height so its final velocity v will be 0 m/s .

Using Kinematic Equation

v = u + gt

On substituting the value we get

➥ 0 = 150 + 10 × t

➥ 150 = 10 × t

➥ 150/10 = t

➥ 15 s = t

So, it will take 15 seconds .

vladimir1956 [14]2 years ago

3 0

Answer:

maximum height reached: 15 s

Explanation:

using the formula: maximum height reached = [(sin(θ) * speed)/ gravity]

Here speed is 150m/s and gravity is 10 m/s² and as it does directly upwards, it has an angle of 90°

using the formula:

maximum height reached = [(sin(90) * 150)/ 10] = 15 s

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The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre

mestny [16]

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

D =150 10⁶ km (1000m / 1 km)

D = 150 10⁹ m.

We must take the given angle to radians.

1º = 3600 arc s

π rad = 180º

θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

θ = x / D

x = θ D

x = 4.85 10-6 150 109

x = 727.5 103 m

x = 727.5 km

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3 years ago

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Marina86 [1]

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A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second

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3 years ago

Read 2 more answers

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$