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olya-2409 [2.1K]

3 years ago

7

Make a claim as to whether analog signals or digital signals are more reliable to transmit information. Use evidence and reasoni

ng to support you claim

1 answer:

cestrela7 [59]3 years ago

3 0

ok this is the answer

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What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What i

inn [45]

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

8 0

3 years ago

Which of the following can a cell contain. A. a number B. a word C. a function D.all of these.

Verizon [17]

A cell can contain D.) all of these

4 0

3 years ago

Read 2 more answers

A flat disk, a solid sphere, and a hollow sphere each have the same mass m and radius r. The three objects are arranged so that

Paladinen [302]

Answer:

$I_{disc} = \frac{1}{2}mR^2$

$I_{sphere} = \frac{2}{5}mR^2$

$I_{hollow} = \frac{2}{3}mR^2$

so largest moment of inertia is for Hollow Sphere

Explanation:

Moment of inertia flat disc is given as

$I_{disc} = \frac{1}{2}mR^2$

moment of inertia of solid sphere

$I_{sphere} = \frac{2}{5}mR^2$

moment of inertia of hollow sphere

$I_{hollow} = \frac{2}{3}mR^2$

all the three objects are given with same mass and same radius

so largest moment of inertia is for Hollow Sphere

8 0

4 years ago

Dana has a sports medal suspended by a long ribbon from her rearview mirror. As she accelerates onto the highway, she notices th

ivann1987 [24]

Answer:

$2.26m/s^2$

Explanation:

Suppose the tension in the ribbon is T.

The upwards component of the tension = Tcos13

The horizontal component of the tension = Tsin13

The weight of the medal = mg

In a vertical direction, mg is balanced by Tcos13, so mg=Tcos13, so T = mg/cos13. The horizontal component of T is the force causing m to accelerate.

$F = ma\\T\sin13^{\circ} = ma$

Substituting T= mg/cos13 gives:

$\frac{mg}{cos13^{\circ}}sin13^{\circ} = ma$

$m$ cancels and you have

$g\tan13 = a\\a=9.81\tan13^{\circ} \approx 2.26m/s^2$

7 0

4 years ago

When a pitcher throws a curve ball, the ball is given a fairly rapid spin?

Oksi-84 [34.3K]

Yes it would be rapid

4 0

4 years ago