The chart shows data for four moving objects.
Object
Initial Velocity
Final Velocity
Change in
(m/s)
(m/s)
Time (s)
29
6
X>N
10
34
12
12
40
20
28
8
Which object has the greatest acceleration?
W
x
Y
Z

Kinematics is the branch of physics and a subdivision of classical mechanics concerned with the geometrically possible motion of a body or system of bodies without consideration of the forces involved. :)

Answer:

Kinematics is a branch of mechanics that deals with pure motion, without reference to the masses or forces involved in it.

Explanation:

Ex. X=2t+1 here x is related to t which is time(function of time)

Answer:

when a medium begins to vibrate

Answer:

I think this is the answer

Explanation:

Answer:

A 1.56    see below

Explanation:

I will assume the 45 degree force is upward

vertical component =   600 sin 45 = 424.26 N

 added to the other vertical force will total 824.26 N

    F = ma

   824.26 = 100 * a    shows a = 8.24 m/s upward

Now we have to assume the mass is ALSO acted on by gravity and this value is given as  9.8  ( so downward is positive)

9.81 -   8.24      =  1.56 m/s^2

We need frequency:-

[tex]\\ \rm\Rrightarrow \nu=\dfrac{c}{\lambda}[/tex]

[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8m/s}{590\times 10^{-9}m}[/tex]

[tex]\\ \rm\Rrightarrow \nu=0.0051\times 10^{17}Hz[/tex]

[tex]\\ \rm\Rrightarrow \nu=5.1\times 10^{14}Hz[/tex]

[tex]~~~~~\dfrac{T_1}{T_2}= \dfrac{1.22}{0.54}\\\\\\\implies \dfrac{2\pi \sqrt{\dfrac{L_1}{g}}}{2\pi \sqrt\dfrac{L_2}g}}= 2.259\\\\\\\implies \dfrac{\sqrt{L_1}}{\sqrt g} \times \dfrac{\sqrt g}{\sqrt{L_2}} = 2.259\\\\\\\implies \sqrt{\dfrac{L_1}{L_2}} = 2.259\\\\\\\implies \dfrac{L_1}{L_2} = 5.104\\\\\\\implies L_1:L_2 = 5.104[/tex]

Answer:

see below

Explanation:

Kind of like when yo open a warm soda .....excess gas is released

Let's see

[tex]\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)[/tex]

Now we know the formula of acceleration

[tex]\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)[/tex]

[tex]\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2[/tex]

[tex]\\ \rm\Rrightarrow \alpha=-x\omega^2[/tex]

Or

[tex]\\ \rm\Rrightarrow \alpha=-\omega^2x[/tex]

Given that the position x of a particle along X-axis varies with time t by the equation:

[tex]{:\implies \quad \sf x=A\sin (\omega t)}[/tex]

As it defines the position, so x is just displacement here, and we need to find the acceleration first for telling with what if varies, so by definition, the second differential coefficient of displacement is acceleration, so differentiating both sides w.r.t.x of the above equation in accordance with chain rule we have:

[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\cos (\omega t)\cdot \omega \cdot \dfrac{dt}{dt}}[/tex]

[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\omega \cos (\omega t)}[/tex]

Differentiating both sides w.r.t.x by chain rule again to get the 2nd order derivative

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega \sin (\omega t)\cdot \omega \dfrac{dt}{dt}}[/tex]

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega^{2}\sin (\omega t)}[/tex]

Re-write as :

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-\omega^{2}\{A\sin (\omega t)\}}[/tex]

Can be further written as

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-x\omega^{2}}[/tex]

This is the Required answer

If they ask you the differential equation for the acceleration of a wave (as the given equation was general equation of a wave), you can simply write:

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}+x\omega^{2}=0}[/tex]

Answer:

The water cycle shows the continuous movement of water within the Earth and atmosphere. It is a complex system that includes many different processes. Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow.

Explanation:

Answer:

30 seconds

Explanation:

The formula for calculating acceleration is

( final velocity - inital velocity) ÷ time

If we enter the values, it would be (250 - 100) ÷ t = 5m/s^2

Now we need to know 't'

So we rearrange the equation to make t the subject.

(250-100) ÷ 5 = 30s

Answer:

7.07 radians per second

Answer~

Both life science and physical science helps to understand the natural habitat of the earth. Life science deals with the living thing; thus, it is a study of the organic world. On the other hand, physical science deals with nonliving things; thus, it is a study of the inorganic world.

Explanation:

For an airplane of mass 270,000 kg , the magnitude of the force exerted is mathematically given as

N2=1584431.13N

N1=1061568.87N

Generally, the equation for the Force balance is mathematically given as

N1+N2=270000*g

Therefore

N2*10=N1*15

N1=10N2/15

N1=0.67N2

Hence

0.67N2+N2=270000*g

N2=1584431.13N

N1=1061568.87N

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Answer:

The quantity and quality of sleep both impact memory.

Explanation:

Answer:

Like a galvanometer, an electric motor contains (a switch, an electromagnet) that is free to rotate between the poles of a permanent, fixed magnet. 9.

The 3rd  overtone or the third harmonic given that the fundamental frequency is 330Hz is 990Hz

Hamonics defined as the number gotten by multiplying the nth overtone by the fundamental frequecency.

In summary, the expression for the nth Hamonic or overtone is given as

Fn = Fundamental Frequency*N

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Answer:

I have no clue I'm just trying to get points

Explanation:

:) sorry

true

Explanation:

It has friction when objects move through it.

Answer:

Here, m=10 kg

The resultant force acting on the body is

F=(98N)2+(6N)2=10N

Let the resultant force F makes an angle θ w.r.t. 8N force.

From figure, tanθ=8N6N=43

The resultant acceleration of the body is 

a=mF=10kg10N=1ms−2

The resulatnt acceleration is along the direction of the resulatnt force.

Hence, the resultant acceleration of the body is 1 ms−2 at an angle of tan−1(43) w.r.t. 8N force.



Explanation:

i dont to i want to copy and paste on my printer

Answer:

The giraffe is the tallest of all mammals. It reaches an overall height of 18 ft (5.5 m) or more. The legs and neck are extremely long. The giraffe has a short body, a tufted tail, a short mane, and short skin-covered horns

Answer:

M g = M a = a (9 - 7) kg      with the two masses going in different directions

(7 + 9) kg * g = a * 2 kg       2 kg effective in prducing acceleration

a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2

M1 a = T - M1 * g        T producing upward acceleration of 1.23 m/s^2

T = 7 (a + g) = 7 * 11 = 77 N

Check:

M2 a =  M2 * g - T          M2 producing downward acceleration

T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N

Answer: The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.

Explanation:

From the activity values and the decay constant, the mass of of Strontium in the sample is:

[tex]1.62 × 10^{-7}g[/tex]

The decay constant of an element is the probability of decay of a nucleus per unit time.

{λ = ln 2 / t1/2 

where;

t1/2 is the half-life of the isotope.

The half-life is converted to seconds since the decay constant is asked in per seconds.

[tex]28.8 years = 28.8 × 3.156 × 10^{7} = 908928000 s \\ [/tex]

Hence;

[tex]λ = \frac{ln2}{90892800s} = 7.626 s^{-1}[/tex]

The activity of the element, A, the decay constant, λ and the number of nuclei, N are related as follows:

[tex]N = \frac{8.25 ×10^{5}}{7.626×10^{-10}} = 1.082 × 10^{15} [/tex]              

Molar mass of Strontium-90 is 90 g.

1 mole of Strontium-90 contains 6.02×10^23 nuclei.

The mass, m of Strontium in the sample is calculated:

[tex]m = 1.082 × 10^{15} × \frac{90 g}{6.02 × 10^{23}} = 1.62 × 10^{-7}g \\ [/tex]

Therefore, the mass of of Strontium in the sample is:

[tex]1.62 × 10^{-7} \: g[/tex]

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Hi there!

In order for a block to begin sliding, the force due to STATIC friction must be overcome.

In this instance, the following forces are acting on the block ALONG the axis of the incline.

Force due to gravity:

On an incline, the component of the force due to gravity contributing to the object's downward movement is equivalent to the horizontal (sine) component.

[tex]F_g = Mgsin\theta[/tex]

Force due to static friction:
The force due to friction is equivalent to the normal force multiplied by the coefficient of friction.

The normal force is the cosine component (perpendicular to the incline), so:
[tex]N = Mgcos\theta\\\\F_s = \mu_sMgcos\theta[/tex]

To find the minimum angle for the block to begin sliding, we can set the two forces equal to 0. They work in opposite directions (let down the incline be negative and up the incline be positive).

[tex]\Sigma F = F_s - F_g\\\\0 = F_s - F_g\\\\0 = \mu_sMgcos\theta - Mgsin\theta\\\\Mgsin\theta = \mu_sMgcos\theta[/tex]

Cancel out 'Mg' and rearrange to solve for theta.

[tex]sin\theta = \mu_scos\theta\\\\tan\theta = \mu_s\\\\\theta = tan^{-1}(\mu_s) = tan^{-1}(.45) = \boxed{24.228^o}[/tex]

Hi there!

For an object on an incline with friction being pulled, the following forces are present.

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
[tex]\Sigma F = F_T + F_f - F_g[/tex]

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
[tex]0 = F_T + F_f - F_g[/tex]

Now, we can express each force as an equation.

Force due to tension:

Force due to gravity:

This is expressed as:
[tex]F_g = Mgsin\theta[/tex]

Force due to friction:

[tex]N = Mgcos\theta[/tex]

[tex]F_f = \mu Mgcos\theta[/tex]

Now, plug these expressions into the above equation.

[tex]0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T[/tex]

Mg = 245 N (weight). Plug in all values:
[tex]245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}[/tex]

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

[tex]N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}[/tex]

To find the tension in the rope needed to make the crate move at a constant velocity, we need to consider the forces acting on the crate.

1. Tension in the rope (T): This force is applied upwards at an angle of 20 degrees above the horizontal.

2. Weight of the crate (W): This force acts downward vertically and is equal to 245 N.

3. Normal force (N): This force is exerted by the floor on the crate and acts perpendicular to the surface of the floor.

4. Kinetic friction force (f_k): This force opposes the motion of the crate and acts parallel to the surface of the floor. The magnitude of kinetic friction is given by: f_k = μ_k * N, where μ_k is the coefficient of kinetic friction (0.270 in this case).

Since the crate is moving at a constant velocity, the net force on the crate must be zero. In other words, the forces pulling the crate forward (T and the horizontal component of the weight) must balance the forces opposing its motion (kinetic friction). The vertical forces (the vertical component of the weight and the normal force) must also balance.

Let's proceed with the calculations:

Horizontal Forces:

T * cos(20°) - f_k = 0

T * sin(20°) + N - W = 0

We can now solve for T and N:

From the horizontal forces equation:

T * cos(20°) = f_k

vertical force:

T = f_k / cos(20°)

T = (0.270 * N) / cos(20°)

From the vertical forces equation:

T * sin(20°) + N = W

T * sin(20°) + N = 245 N

Now, substitute the expression for T from the horizontal forces equation:

(0.270 * N) / cos(20°) * sin(20°) + N = 245 N

Now, solve for N:

N * (0.270 * tan(20°) + 1) = 245 N

N = 245 N / (0.270 * tan(20°) + 1)

Now, plug in the values and calculate N:

N = 245 N / (0.270 * tan(20°) + 1)

N ≈ 208.75 N

Now, we can calculate the tension in the rope (T):

T = (0.270 * N) / cos(20°)

T ≈ (0.270 * 208.75 N) / cos(20°)

T ≈ 67.24 N

So, the tension in the rope needed to make the crate move at a constant velocity is approximately 67.24 N, and the normal force exerted by the floor on the crate is approximately 208.75 N.

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The boat will swing back and forth about the same location, never reaching the land. Due to which the boaters plan will not help him reach shore.

The ship rocks back and forth as the wind and waves press against it. Similar to swiveling on a chair, yaw spins the ship on an unseen central line.

Waves travelling perpendicular to the ship's motion might create this, changing the ship's heading or direction.

The motion of the motor will be identical to that of water molecules. Because there is a wave on the water surface,The motor will move in lockstep with the water particle.

The particles of the medium only vibrate perpendicular to the wave propagation as the wave travels through the water surface, but they are not moved from their initial position.

As a result, the boat will swing back and forth about the same location, never reaching the land.

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