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3 years ago

The number ocean waves that pass a buoy in one second is _ of the wave

2 answers:

7 0

The number of ocean waves that pass a buoy in one second is the frequency of the wave. The crest of a transverse wave is its highest point.

alexgriva [62]3 years ago

4 0

Your answer is the frequency of the wave

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It’s 1 &2 question ! please zoom in to see it clearly :)

yarga [219]

Answer:

1. C

2. D

Explanation:

1. as the object is stationary despite Jamal applying 100N forces, it suggests that the object is applying an equal amount of force on the opposite direction that is opposing the force applied by Jamal.

2. inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.

hope it helps. :)

5 0

3 years ago

Object A weighs 750 N on earth. Object B weighs 750 N on Jupiter.

Katarina [22]

The object A has the greater mass compared to object B.

Explanation: 

The weight of any object on any planet is the measurement of gravity’s influence acting on the mass of the object. So for Earth, the acceleration will be acting on the object A’s mass (m) in Earth leading to the weight of the object A as 750 N.

While the acceleration of Jupiter will be acting on the object B’s mass kept in Jupiter to attain the weight of 750 N. So, the mass of both the objects at their respective planet will vary depending on the acceleration of each planet. We can check this as below:

$\text { object A's weight }=\text { m of object A } \times \text { Acceleration of Earth}$

So,

$750 \mathrm{N}=\text { m of object } A \times 9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus,

$\text { m of object } A=\frac{750}{9.8}=76.5 \mathrm{kg}$

Similarly for object B,

$\text { Weight of object } B=m \text { of object } B \times \text { Acceleration due to gravity of Jupiter }$

$750 = m \text { of object } B \times 24.79 \mathrm{m} / \mathrm{s}^{2}$

Thus,

$\text { m of the object } B=\frac{750}{24.79}=30.25 \mathrm{kg}$

Thus, the mass of object A is greater than the mass of object B.

5 0

3 years ago

charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel

IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁ + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³ N / C .

3 0

3 years ago

Question 2 of 10

Arlecino [84]

Answer:

B: Energy that is transformed is neither created or destroyed

Explanation:

4 0

2 years ago

1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of

DiKsa [7]

Answer:

44.6 N

Explanation:

Draw a free body diagram of the block. There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° − mg = 0

N = mg − F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° − Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg − F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N

8 0

3 years ago

Read 2 more answers