A device that measures potential is a(n) circuit.

Which question cannot be answered through making measurements?

Bond [772]

A. is the answer for this question

5 0

3 years ago

A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3

Klio2033 [76]

Answer:

$\theta=30^{\circ}$

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

$F=ILB\ sin\theta$

$sin\theta=\dfrac{F}{ILB}$

$sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}$

$\theta=30^{\circ}$

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0

3 years ago

A train moves from rest to a speed of 25 m/s in 30.0 seconds. What is the acceleration?

fredd [130]

Answer:

$a = 0.83\ m/s^2$

Explanation:

<u>Uniform Acceleration </u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting:

$\displaystyle a=\frac{25-0}{30}$

$\boxed{a = 0.83\ m/s^2}$

4 0

3 years ago

In this activity, you will research how scientists developed historical

Darya [45]

Answer: Bohr's model (1913)

Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles.

Bohr's model (1913)

Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

John Dalton

The idea that everything is made of atoms was pioneered by John Dalton (1766-1844) in a book he published in 1808. He is sometimes called the "father" of atomic theory, but judging from this photo on the right "grandfather" might be a better term.

Explanation:

8 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$