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4 years ago

Why can’t you use a model to learn everything about an object?

Physics

2 answers:

kogti [31]4 years ago

4 0

Answer:

Models have always been important in science and continue to be used to test hypotheses and predict information. Often they are not accurate because the scientists may not have all the data. It is important that scientists test their models and be willing to improve them as new data comes to light.

Explanation:

goldenfox [79]4 years ago

3 0

Answer:Models have always been important in science and continue to be used to test hypotheses and predict information. Often they are not accurate because the scientists may not have all the data. It is important that scientists test their models and be willing to improve them as new data comes to light.

Explanation:

You might be interested in

The California sea lion is capable of making extremely fast, tight turns while swimming underwater. In one study, scientists obs

anygoal [31]

Answer:

Acceleration of Sea Lion is 4.41 g

This is 49% of maximum jet acceleration given as a = 9g

Explanation:

As we know that the radius of the circular loop is given as

R = 0.37 m

The speed of the fish is given as

$v = 4 m/s$

Now the centripetal acceleration of the sea lion is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{0.37}$

$a_c = 43.2 m/s^2$

as we know that

$g = 9.8 m/s^2$

so we have

$a = 4.41 g$

Now Percentage of this acceleration wrt maximum jet acceleration is given as

$P = \frac{4.41 g}{9g} \times 100$

$P = 49$ %

6 0

3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago

A sound wave has a speed of 343 m/s in air. What is the wavelength of a sound wave with a frequency of 686 Hz

Svetradugi [14.3K]

Wavelength is the distance between 2 adjacent points in a wave

we can use the following equation to find the wavelength of a sound wave

wavelength = speed / frequency

frequency is the number of waves passing a point in 1 second

substituting the values in the equation

wavelength = 343 m/s / 686 Hz

wavelength = 0.5 m

wavelength of the wave is 0.5 m

6 0

3 years ago

a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':

Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

$v=\sqrt{u^{2}+2as}$

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

$v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s$

Also, v=u+at and making t the subject of the formula

$t=\frac {v-u}{a}$

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

$t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s$

Therefore, it needs 7.6 seconds to travel

7 0

3 years ago

What are two ways thermal energy can be increased in a system?

MaRussiya [10]

Adding thermal energy
Performing work on the system

3 0

3 years ago