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3 years ago

HELLPPP NOWWA!!!!!!!!!!!!

Physics

2 answers:

elena55 [62]3 years ago

8 0

Answer:

it would be a

Explanation:

i remeber taking the test im pretty sure its a

prisoha [69]3 years ago

7 0

Answer:

Explanation:

the gravitational pull also helps with that but

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Who is known as the "father of modern chemistry" because he first organized all known elements into four different

iVinArrow [24]

Answer:

D. Henry Moseley

Explanation:

I'm chemistry right now

5 0

3 years ago

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A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a relation between the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

6 0

3 years ago

A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p

MA_775_DIABLO [31]

There are some placeholders in the expression, but they can be safely assumed

Answer:

(a) $f=1617.9\ Hz$

(b) $T=0.618\ ms$

(d) $\varphi=60^o$

Explanation:

Sinusoidal Waves

An oscillating wave can be expressed as a sinusoidal function as follows

$V(t)&=A\cdot \sin(2\pi ft+\varphi )$

Where

$A=Amplitude$

$f=frequency$

$\varphi=Phase\ angle$

The voltage of the question is the sinusoid expression

$V(t)=20cos(5\pi\times 103t+60^o)$

(a) By comparing with the general formula we have

$f=5\pi\times 103=1617.9\ Hz$

$\boxed{f=1617.9\ Hz}$

(b) The period is the reciprocal of the frequency:

$\displaystyle T=\frac{1}{f}$

$\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec$

Converting to milliseconds

$\boxed{T=0.618\ ms}$

$\boxed{A=20 \ Volts}$

(d) Phase angle:

$\boxed{\varphi=60^o}$

4 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

4 years ago

Consider Newton's Law of Universal Gravitation: FG= G (m1 m2)/d2 .

bixtya [17]

Answer:

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

Explanation:

The formula for the force of gravity between two objects is

$F=G\frac{m_1 m_2}{d^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

d is the separation between the two objects

We notice the following:

- F is directly proportional to the masses, $F\propto m_1, m_2$ . This means that if one of the masses increases, then the force between them, F, increases in a proportional way

- F is inversely proportional to the square of the distance, $F\propto \frac{1}{d^2}$ . This means that if the distance between the two objects is increased, the force between them will decrease, and vice-versa.

So, the correct answer is

7 0

3 years ago

Read 2 more answers