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3 years ago

Could anyone solve this...its kinda urgent.

Physics

1 answer:

LenKa [72]3 years ago

5 0

Answer:

1) A = lost volts, B = lower, C = a small pressure, D, = Newtons kg, E = Modulation.

2) A = True(?), B = true, C = True, D = True, E = not sure sorry.

Explanation:

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A boat sails south with the help of a wind blowing in the direction s36°e with magnitude 300 lb. find the work done by the wind

olga nikolaevna [1]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
<span>Work=effective force x distance = 300cos36. 100 ft.lb.</span>

6 0

3 years ago

The Sears Tower is nearly 400 m high. How long would it take a steel ball to reach the ground if dropped on the top? What will b

kipiarov [429]

Answers:

a) 9.035 s

b) -88.543 m/s

Explanation:

The described situation is related to vertical motion (especifically free fall) and the equations that will be useful are:

$y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}$ (1)

$V=V_{o}+gt$ (2)

Where:

$y=0$ is the final height of the steel ball

$y_{o}=400 m$ is the initial height of the steel ball

$V_{o}=0$ is the initial velocity of the steel ball (it was dropped)

$V$ is the final velocity of the steel ball

$t$ is the time it takes to the steel ball to reach the ground

$g=-9.8 m/s^{2}$ is the acceleration due to gravity

<u>Knowing this, let's begin with the answers:</u>

<h2>a) Time it takes the steel ball to reach the ground</h2>

We will use equation (1) with the conditions listed above:

$0=y_{o}+\frac{1}{2}gt^{2}$ (3)

Isolating $t$ :

$t=\sqrt{\frac{-2y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2(400 m)}{-9.8 m/s^{2}}}$ (5)

$t=9.035 s$ (6)

<h2>b) Final velocity of the steel ball</h2>

We will use equation (2) with the conditions explained above and the calculaated time:

$V=gt$ (7)

$V=(-9.8 m/s^{2})(9.035 s)$ (8)

$V=-88.543 m/s$ (9) The negative sign indicates the direction of the velocity is downwards

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3 years ago

Ty is three years old and as a result of a “stomach bug" has been vomiting for

nataly862011 [7]

Answer:

Ty's blood is slightly alkaline

Explanation:

3 0

2 years ago

How many O atoms are in 8 molecules of C²O⁴H

MariettaO [177]

Answer:

32 oxygen atoms

6 0

3 years ago

A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at

zhuklara [117]

Answer:

$B=1.21*10^{-10}T$

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

$B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)$

$\mu_0$ is the permeability of free space, $I$ is the flowing current through the ring, $R$ is the ring's radius and $r$ is the distance to the center of the ring.

The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period $T=\frac{2\pi}{\omega}$ . So, we have:

$I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A$

Now, replacing in (1):

$B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T$

6 0

4 years ago

Could anyone solve this...its kinda urgent.​

Could anyone solve this...its kinda urgent.