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3 years ago

a cell phone uses 3.0V battery. the circuit board needs a 0.05 A current. what size resistor is needed to generate this current

Physics

1 answer:

san4es73 [151]3 years ago

7 0

Answer:

60 Ohms

Explanation:

Ohms law states that the voltage in the circuit is directly proportional to the current through the circuit components and expressed as

V=IR

Where V is the voltage, I is current and R is resistance

Making R the subject of the formula then

$R=\frac {V}{I}$

Substituting 3.0V for V and 0.05 A for I then

$R=\frac {3}{0.05}=60.0$

Therefore, resistance is 60.0 Ohms

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As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe

Sidana [21]

Answer:

It remains constant

Explanation:

As we know that buoyant force on an object given as

Fb = ρ Vd g

ρ= Density of fluid

Vd=Volume displace by body

g=10 m/s²

Fb =buoyant force

So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.

So when rock gets deeper and deeper the buoyant force will remain constant.

It remains constant

3 0

3 years ago

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

Bas_tet [7]

Answer:

W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

F = G m₁ M / r²

W = ∫ F. dr

W = G m₁ M ∫ dr / r²

we integrate

W = G m₁ M (-1 / r)

We evaluate between the limits, lower r = R_ Moon and r = ∞

W = -G m₁ M (1 /∞ - 1 / R_moon)

W = G m1 M / r_moon

Body weight is

W = mg

m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

m = 1000/32 = 165 / g_moon

g_moom = 165 32/1000

.g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system

W_capsule = 1000 pound (1 kg / 2.20 pounds)

W_capsule = 454 N

W = m_capsule g

m_capsule = W / g

m = 454 /9.8

m_capsule = 46,327 kg

Let's calculate

W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶

W = 1,307 10⁶ J

7 0

3 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction

4 0

3 years ago

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

Sidana [21]

Answer:

255.4 N/m

Explanation:

We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where in this case, we know:

f = 29 Hz is the frequency of oscillation

k is the spring constant, which is unknown

m = 7.7 g = 0.0077 kg is the mass of the eyeball

Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:

$k=(2\pi f)^2 m=(2 \pi (29 Hz))^2 (0.0077 kg)=255.4 N/m$

6 0

3 years ago

An object is placed 8.5 cm in front of a convex spherical mirror of focal length −14.0 cm. What is the image distance? Show all

Pani-rosa [81]

Given:

u = 8.5 cm

f = -14.0 cm

v = image distance

Using the mirror formula 1/u + 1/v = 1/f

1/8.5 + 1/v = 1/-14.0

Rewrite to solve for v:

v = (8.5 * -14.0) / (8.5 - (-14.0))

v = -119 / 22.5

v = -5.29 cm ( round answer as needed.)

5 0

3 years ago