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ahrayia [7]
3 years ago
13

Laptop computers are made with batteries, but they must also be plugged into outlets to charge the batteries. which is true rega

rding laptops? they run only on alternating current. they run only on direct current. they can run on either direct or alternating current. they do not run on direct or alternating current.
Physics
1 answer:
kiruha [24]3 years ago
4 0

The true statemement regarding laptops is they can run on either direct or alternating current.

<h3>What is direct current source?</h3>

This is the type of current that flows only in one direction. The source of this type of current in battery.

<h3>What is alternate current source?</h3>

This is the type of current that changes direction with time. The source of this type of current is current from outlet.

Thus, the true statemement regarding laptops is they can run on either direct or alternating current.

Learn more about direct current here: brainly.com/question/10605331

#SPJ1

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As the wavelength increases, the frequency? A. decreases and energy decreases B. increases and energy increases C. decreases and
const2013 [10]

For the same wave, the product product of

               (wavelength) times (frequency)

is always the same number.  (It happens to be the speed of the wave.)

So if one of them changes, the other one has to change in the opposite
direction, in order to keep their product constant.

For electromagnetic waves, higher frequency means higher energy.
I'm not sure about mechanical waves just now.

8 0
4 years ago
I don’t understand the 4th one please help someone.
Gnoma [55]

Answer:

From point A to point D is 20

The final displacement is 32

Explanation:

AB=8

BC=8

CD=4

DE=8

EF=4

DE= 8 because the object moves 4 meters in each direction.

AB+BC+CD=20(A to D)

AB+BC+CD+DE+EF=32(Final displacement)

7 0
3 years ago
An archer wishes to shoot an arrow at a target at eye level a distance of 50.0m away. If the initial speed imparted to the arrow
notka56 [123]
<span>Given:
Hmax (distance) = 50.0m
v</span>₀ = <span>70.0m/s

Required:
what angle should the arrow make with the horizontal as it is being shot

Solution:
Hmax = v</span>₀²sin²θ / 2g
sin²θ = 2gHmax / v₀²
sin²θ = 2 (9.81 m/s²) (50m) / (70 m/s)²
sin²θ = 0.200
θ = 26.56°
5 0
3 years ago
I’M DESPERATE PLZ HELP!! 40 POINTS!!
Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, F_k, required is given by the formula, F = m × a,

From which we have;

F_k = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, F_k = 0.0625 N

c. The coefficient of friction between the ice and puck, \mu_k, is given from the equation for the kinetic friction force as follows;

F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8

\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755

The coefficient of friction between the ice and puck, \mu_k ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force F_k = 0.012755 \times 1 \times 9.8 = 0.125  \ N

The acceleration, a = F_k/m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here,  F_k  = 0.125  \ N

The coefficient of friction \mu_k ≈ 0.012755 is constant

5 0
3 years ago
Suppose a spring has a relaxed length of 25.1 cm. The simulation refers to this as the natural length. This is the length of the
Leviafan [203]
The anwser is 30096 hope it helps bud
7 0
4 years ago
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