Answer ASAP and only if you know its correct

Nuclear Size and Mass. The radii of most nuclei is given by the equation: R = R. 0. A1/3. R: radius of the nucleus, A: mass numb

mina [271]

Answer:

sssss

Explanation:

Hope this helps you

3 0

2 years ago

Good morning friends

Hatshy [7]

Answer:

Good morning. Whats about you

4 0

2 years ago

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Examples of reaction force and action force hewlp

mrs_skeptik [129]

Answer:

Action-Reaction Force Examples in Everyday Life

Recoil of a Gun.

Swimming.

Pushing the Wall.

Diving off a Raft.

Space Shuttle.

Explanation:

hope this helps

6 0

2 years ago

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The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the

Ivanshal [37]

Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

(c) $\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) $a = 22.458\ m/s^{2}$

Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

$omega'$ = final angular velocity

$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

$\alpha = - 1.32\ rev/m^{2}$

Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

$\theta = 13674\ rev$

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

$\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) The radial acceleration is given by:

$\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s$

Linear acceleration is given by:

$a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}$

$a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}$

5 0

2 years ago

You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Ap

Inga [223]

The answer is 35 minutes

The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ

T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant

step 1. Calculate constant k:

We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?

T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ*¹⁰
200 = 68 + 142 * e⁻¹⁰ⁿ
200 - 68 = 142 * e⁻¹⁰ⁿ
132 = 142 * e⁻¹⁰ⁿ
e⁻¹⁰ⁿ = 132/142
e⁻¹⁰ⁿ = 0.93

Logarithm both sides with natural logarithm:
ln(e⁻¹⁰ⁿ) = ln(0.93)
-10n * ln(e) = -0.07
-10n * 1 = - 0.07
-10n = -0.07
n = -0.07 / - 10
n = 0.007

Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
Ta = 68°F
To = 210°F
n = 0.007

T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79

Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = ln(0.79)
-0.007x * ln(e) = -0.24
-0.007x * 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35

5 0

2 years ago

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