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umka21 [38]
3 years ago
8

An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m

and has an initial velocity of v0 = 2.2 m/s. How far will the object move before it achieves a velocity of v = 7.2 m/s? Your answer should be accurate to the nearest 0.1 m.
Physics
1 answer:
solong [7]3 years ago
8 0

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

= 3.9m + 3.6m = 7.5m

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Answer:

Squids = 450 - 490 nm (Moderate Frequency) (Blue)

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Frogs = 280 - 580 nm (Very Low Frequency)

Explanation:

All of the above mentioned ranges are compared to that of humans.

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3 0
3 years ago
What is the source of all mechanical waves?
frutty [35]

Answer:

The correct answer is A. Vibration.

Explanation:

Mechanical waves is formed by the oscillation of matter and therefore transfer energy from one medium to the other. Unlike electromagnetic waves, mechanical waves need some medium to propagate. It requires an initial energy input and thus carries this energy when it propagates. There are three types of mechanical waves namely transverse waves, longitudinal waves and surface waves. Examples of such waves are sound waves, water waves and seismic waves.

6 0
3 years ago
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
A jumbo jet weighing 65,000 kg Is accelerating at 3 m/s^2 while its jet engines are putting out 200000 N of thrust.
sveticcg [70]

Answer:

Fa = 5000 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

Let's assume that the movement of the plane is to the right, any movement or force to the right will be marked with a positive sign, while any force or movement to the left, will be taken as negative.

The force of the turbine drives the plane to the right, therefore it is positive, the acceleration is constant and keeps the movement to the right, therefore it is positive, the wind drag force tries to prevent the movement of the plane to the left therefore it is negative, with this analysis we deduce the following equation.

ΣF = m*a

where:

ΣF = sum of forces [N] (units of Newtons)

m = mass = 65000 [kg]

a = acceleration = 3 [m/s²]

Fa = force exerted by the air [N]

200000 - Fa = 65000*3

Fa = 200000 - (3*65000)

Fa = 5000 [N]

8 0
3 years ago
Greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
maw [93]

{\large{\bold{\rm{\underline{Given \; that}}}}}

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

{\large{\bold{\rm{\underline{To\; find}}}}}

★ The speed of the hound and the hare

{\large{\bold{\rm{\underline{Solution}}}}}

★ The speed of the hound and the hare = 25:18

{\large{\bold{\rm{\underline{Full \; Solution}}}}}

\dashrightarrow  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

 But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

 Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

  • (5/3)a × 5

  • 25/3a metres

 Now the distance travelled by the hare in it's 6 leaps..!

  • 6a metres

 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

  • 25/3a = 6a

  • 25/3 = 6

  • 25:18
5 0
3 years ago
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