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3 years ago

PLEASE HELP ME I NEED AN ANSWER BY 10 PM OR I WILL GET AN F PLLLLLLLEEEEEEEEEEEEAAAAAAAAAAAASSSSSSSSSSSSSEEEEEEEEEEE

Physics

2 answers:

jolli1 [7]3 years ago

8 0

Here is a example of Newton

masya89 [10]3 years ago

3 0

Newton's 2nd law of motion: Force = (mass) x (acceleration)

You need to find the acceleration, so divide each side by (mass):

Acceleration = (Force) / (mass)

Plug in the numbers you know from the question:

Acceleration = (300,000 Newtons) / (20,000 kilograms)

Acceleration = 15 m/s²

This is a huge acceleration. It's 53% faster than a falling rock (gravity).

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What type of blood does the night side of the heart receive?

lozanna [386]

Answer:

The answer is 92.852 million mi.

Explanation:

6 0

3 years ago

soaring birds and glider pilots can remain aloft for hours without expending power. Discuss why this is so.

vekshin1

Answer:

Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection

Explanation:

5 0

3 years ago

2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗

Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The tension is $T = 48.255 \ N$

The time taken is $t = 0.226 \ s$

The position for maximum velocity is

S = 0

The maximum velocity is $V_{max} =0.384 \ m/s$

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

The mass of the bob is $m_b = 5 \ kg$

The angle is $\theta = 10^o$

The length of the string is $L = 0.5 \ m$

The tension on the string is mathematically represented as

$T = mg cos \theta$

substituting values

$T = 5 * 9.8 cos(10)$

$T = 48.255 \ N$

The motion of the bob is mathematically represented as

$S = A sin (w t + \frac{\pi }{2} )$

=> $S = A sin (wt)$

Where $w$ is the angular speed

and $\frac{\pi}{2}$ is the phase change

At initial position S = 0

So $wt = cos^{-1} (0)$

$wt = 1$

Generally $w$ can be mathematically represented as

$w = \frac{2 \pi }{T}$

Where T is the period of oscillation which i mathematically represented as

$T = 2 \pi \sqrt{\frac{L}{g} }$

$t = \frac{1}{w}$

$t = \frac{T}{2 \pi}$

$t = \sqrt{\frac{L}{g} }$

substituting values

$t = \sqrt{\frac{0.5}{9.8} }$

$t = 0.226 \ s$

Looking at the equation

$wt = 1$

We see that maximum velocity of the bob will be at S = 0

i. e $w = \frac{1}{t}$

The maximum velocity is mathematically represented as

$V_{max} = w A$

Where A is the amplitude which is mathematically represented as

$A = L sin \theta$

$V_{max} = \frac{2 \pi }{T } L sin \theta$

$V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} } L sin \theta$ Recall $T = 2 \pi \sqrt{\frac{L}{g} }$

$V_{max} = \sqrt{gL} sin \theta$

substituting values

$V_{max} = \sqrt{9.8 * 0.5 } sin (10)$

$V_{max} =0.384 \ m/s$

6 0

3 years ago

Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a

Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

P = (m1 + m2) a

a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block. In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

N = m2 a

fr -W2 = 0

fr = W2

The definition of friction force is

fr = μ N

Let's replace and calculate

μ (m2 a) = m2 g

μ (P / (m1 + m2)) = g

P = g /μ (m1 + m2)

Let's calculate the value of this force

P = 9.8 / 0.710 (28.9 +4.4)

P = 13.80 (33.3)

P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0

3 years ago

What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not

mr_godi [17]

Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

$F=G*\frac{m1*m2}{r^{2} }$

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

$F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{2^{2} } }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{4} }$

$F'=4*G*\frac{m1*m2}{r^{2} }$

F'=4*F

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

7 0

3 years ago