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denis-greek [22]
3 years ago
9

Reduce 2.525 liters ti microliters

Physics
1 answer:
wlad13 [49]3 years ago
7 0

Answer:

2525000  microliters

Explanation:

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When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.86 cm. (a) W
Goryan [66]

Answer:

a)  k = 891.82 N/m

b) e = 0.0143 m = 1.43 cm

c) W = 5.02 J

Explanation:

Step 1: Data given

Mass = 2.60 kg

the spring stretches 2.86 cm = 0.0286

Step 2: What is the force constant of the spring?

Force constant, k = force applied / extension produced  

k = (2.60kg * 9.81N/kg) / 0.0326 m

k = 891.82 N/m

b)  If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it

Extension = F/k = (1.30 kg * 9.81) / 891.82 =  0.0143 m = 1.43 cm

Half the mass means half the extension

c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?

W = average force used * distance

W = 1/2 * k*e * e = 1/2 k*e²  

W = 1/2 * 891.82 * (0.075)² = W = 5.02 J

5 0
3 years ago
g Let the orbital radius of a planet be R and let the orbital period of the planet be T. What quantity is constant for all plane
Wittaler [7]

Explanation:

Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

Here,

G is gravitational constant

M is mass of sun

It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.

7 0
3 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, v_t = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

v_t = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, v_t = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0
3 years ago
A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th
fiasKO [112]
Let's use the mirror equation to solve the problem:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}
where f is the focal length of the mirror, d_o the distance of the object from the mirror, and d_i the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for d_i by using the numbers given in the text of the problem:
\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}
\frac{1}{d_i}= -\frac{7}{60 cm}
d_i = -8.6 cm
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
6 0
3 years ago
Read 2 more answers
A 2200 kg car moving east at 10.3 m/s collides with a 3160 kg car moving east. The carsstick together and move east as a unit af
IgorLugansk [536]

Answer:

u = 1.77 m/s

Explanation:

Conservation of momentum

2250(10.3) + 3160u = (2250 + 3160)(5.32)

u = 1.77 m/s

8 0
3 years ago
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