The statement "if I raise the temperature of
a cup of water, then an increased amount of
sugar will be dissolved" is an example of
A. an observation
B. a hypothesis.
C. a theory

Answer:

2000 N.

Explanation:

From the question given above, the following data were obtained:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

The mechanical advantage of a machine is simply defined as:

Mechanical advantage = Load / Effort

MA = L / E

With the above formula, we can obtain the effort used to raise the load of 8000 N as follow:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

MA = L / E

4 = 8000 / E

Cross multiply

4 × E = 8000

Divide both side by 4

E = 8000 / 4

E = 2000 N

Thus, the effort used to raised the load is 2000 N.

Answer:

 v = 4.8 10⁻⁴ m / s

Explanation:

To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy

          U = K

          q V = ½ m v²

          v = [tex]\sqrt { \frac{2qV}{m} }[/tex]

in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C

let's calculate

          v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]

          v = 4.8 10⁻⁴ m / s

Answer:

Instantaneous velocity  [tex]= 20[/tex] meter per second

Instantaneous acceleration  [tex]= 6[/tex] meter per second square

Explanation:

Given equation of distance X = [tex]2t+3t^2[/tex]

Instantaneous velocity [tex]= \frac{dX}{dt}[/tex]  [tex]= 2 + 6 t[/tex]

Substituting the value of t = 3 seconds, we get -

[tex]\frac{dX}{dt} = 2 + 6*3 = 20[/tex] meter per second

Instantaneous acceleration  [tex]= \frac{d^2X}{dt^2}[/tex]  [tex]= 6[/tex] meter per second square

Answer:

[tex]6.28\times 10^{-5}\ \text{T}[/tex]

[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]

[tex]I_l[/tex] = Current in circular loop = 13 A

[tex]r_l[/tex] = Radius of circular loop = 13 cm

[tex]N[/tex] = Number of turns = 58

[tex]r_c[/tex] = Radius of coil = 0.94 cm

[tex]I_c[/tex] = Current in coil = 1.9 A

[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]

Magnitude of magnetic field in circular loop

[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]

The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].

Torque is given by

[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]

The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].

Answer:

A.

Explanation:

Some of these photons are absorbed by the atoms in the ground state and the photons are lost to the laser process. However, some photons cause stimulated emission in excited-state atoms, releasing another coherent photon. In effect, this results in optical amplification

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

Answer:

C

Explanation:

if you measure your body's flexibility then you can keep track of how flexible you have gotten over time

Answer:

100 milliliters

Explanation:

cold, dry, and stable air masses .

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have ―

We've been asked to calculate acceleration.

[tex]\qquad \implies\boxed{\red{\sf{ a = \dfrac{v-u}{t} }}}\\[/tex]

[tex] \twoheadrightarrow \quad \sf {a = \dfrac{30-18}{3} \; ms^{-2} } \\ [/tex]

[tex] \twoheadrightarrow \quad \sf {a =\cancel{ \dfrac{12}{3} \; ms^{-2} }} \\ [/tex]

[tex]\twoheadrightarrow \quad \boxed{\red{\sf{ a = 4 \; ms^{-2} }}}\\[/tex]

Therefore, acceleration of the car is 4 m/s².