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2 years ago

Question 4

Physics

1 answer:

saveliy_v [14]2 years ago

5 0

Answer:

It is the tendency

of an object to resist any change in its state of motion .

Explanation:

if I am right mark my answer as brainliest

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Which of the following is true of the atom below:

Brilliant_brown [7]

Its B. I know because I just took the test in K12 so your welcome EJ.

7 0

3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers

Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers

Dmitrij [34]

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

6 0

4 years ago

As additional resistors are connected in series to a constant voltage source, how is the power supplied by the source affected?

loris [4]

Resistors decrease the power of the source

3 0

3 years ago

Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x--1 m and has a surface charge density

MariettaO [177]

Answer:

Ea = 3.2 10⁵ N / C

Explanation:

To calculate the electric field of each plane we will use Gauss's law, we create a Gaussian surface that is a cylinder that has the axis perpendicular to the plane, in this case the flow line between the cylinder walls and the surface is zero and all the flow is perpendicular to the base of the cylinder.

We apply the law of gauss flow to each side is the value of the electric field (E) for the area of the cylinder (A); whereby the flow in the two directions is 2 E A

Φ = 2E A = ρₙt / εo

Where ρₙ is the charge inside the cylinder, as the charge density gives us, σ = Q / A

ρₙ = σ A

By which we can clear the electric field

E = σ A / 2εo

where it is worth 8.85 10⁻¹² C² / N² m². Let's calculate with this equation in the field for each plane

1 plane σ= -2.0 pC / m²

E1 = -2. 10⁻¹² / 2 8.85 10⁻¹²

E1 = -0.113 N / C

The field line is directed to the plane

2 Plane σ = 5.8 mC / m2

E2 = 5.8 10-6 / 2 8.85 10-12

E2 = 3,277 10 5 N / A

Field lines leave the plane

As we have the values of each field in the whole space. Let's calculate in the field at the point x = 1 m

To do this we must add the fields at the selected point vectorally, for this distance the point is between the two planes, so the field of plane 1 points to the left and the point of plane 2 also points to the left, consequently field adds

Ea = E1 + E2

Ea = 0.113 + 3.277 10⁵

Ea = 3.2 10⁵ N / C

At point X = -1 m in this case the point is on plane 1, so this plane does not generate any field, it is an equipotential surface, the total field is equal to field 2

Ea = E1 = 3.2 10⁵ N / C

Note that there is no difference in numbers values by the difference between the load between each plane

5 0

4 years ago