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irina1246 [14]
2 years ago
11

RADIOWAVES of constant amplitude can be generated with

Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

Answer:

4) oscillator

Explanation:

4) oscillator

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A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is
andre [41]

Answer:

2.80N/m

Explanation:

Given data

mass m= 56kg

perios T= 11.2s

The expression for the period is given as

T=2π√m/k

Substitute

11.2= 2*3.142*√56/k

square both sides

11.2^2= 2*3.142*56/k

125.44= 351.904/k

k=351.904/125.44

k= 2.80N/m

Hence the spring constant is 2.80N/m

8 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
A train increase its speed steadily from 10ms to 20ms in 1 minutes
maxonik [38]

Answer:

Explanation:  nigerian what

8 0
3 years ago
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction
Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed  = 25.5m/s

Acceleration  = 1.94m/s²

Time  = 2.3s

Unknown:

Final speed of the car  = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

    v = u  + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

 Now insert the parameters and solve;

      v  = 25.5 + (1.94 x 2.3)  = 29.96m/s

3 0
3 years ago
Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

\theta=tan^{-1}(0.5)

\theta= 26.56°

Polar coordinates for point C is (3\sqrt{5}, 26.56°)

3 0
3 years ago
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