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4 years ago

12

ingrid kicks a football with an initial velocity of 12 m/s at an angle 45 relative to the ground. what is the horizontal compone

nt of the initial velocity to the nearest tenth

2 answers:

snow_lady [41]4 years ago

8 0

is correct answer => 8.5 m/s

nydimaria [60]4 years ago

4 0

Let's assume that the coordinate system is defined as positive in the first quadrant.

The system has its origin in the place where Ingrid kicks the football ball.

We have then that the horizontal component is given by:

$x = 12cos (45)$

Rewriting we have:

$x = 12 ( \frac{\sqrt{2}}{2})\\x = 8.485281374$

Rounding the obtained result we have:

$x = 8.5 \frac{m}{s}$

Answer:

The horizontal component of the initial velocity to the nearest tenth is:

$x = 8.5 \frac{m}{s}$

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A clock moves past you at a speed of 0.9 c. How much time passes for you for each second that elapses on the moving clock?

Zigmanuir [339]

Answer:

Explanation:

Time dilation formula is

T = T₀ / √ 1-v²/c²

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Here T₀ = 1 second

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So 2.3 second will pass for each second on moving reference.

6 0

3 years ago

You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el

Lapatulllka [165]

Answer:

$a=0.5418\ m.s^{-2}$ upwards

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Explanation:

Given:

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So, the mass of the person:

$m=\frac{w}{g}$

$m=\frac{688}{9.81}$

$m=70.132\ kg$

Now if the apparent weight in the elevator, $w_a= 726\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w_a-w$

$\Delta w=726-688$

$\Delta w=38\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{38}{70.132}$

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Now if the apparent weight in the elevator, $w_a= 598\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w-w_a$

$\Delta w=688-598$

$\Delta w=90\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

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7 0

4 years ago

Read 2 more answers

The Type K thermocouple has a sensitivity of about 41 uV /°C, i.e. for each degree difference in the junction temperature, the o

Novosadov [1.4K]

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of $2^{16}=65536$ steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

$\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C$

3 0

3 years ago

miskamm [114]

B . I hope this is right

4 0

3 years ago

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HACTEHA [7]

Movement can be without change in positions. As only a part of it can be made to move. If you move there is a motion and if there is a motion there is a movement so they're same thing. In motion whole of body moves, but in movements only the parts of body moves.

7 0

3 years ago