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3 years ago

8

Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 5.5 mi2/hr. How rapidly is rad

ius of the spill increasing when the area is 6 mi2?

1 answer:

Travka [436]3 years ago

8 0

<h2>The radius will increase at the rate of 0.64 mi/hr </h2>

Explanation:

The area of a circle can be represented by A = π r² I

Differentiating both sides w.r.t time

$\frac{dA}{dt}$ = 2π r $\frac{dr}{dt}$ II

Dividing II by I , we have

$\frac{dA}{A}$ = 2 x $\frac{dr}{r}$

substituting the values

$\frac{dr}{r}$ = $\frac{5.5}{12}$ = 0.46 mi per unit radius

or dr = 1.4 x 0.46 = 0.64 mi/hr

here 1.4 mi is the radius , when area of circle is 6 mi²

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In a lab, four balls have the same velocities but different masses.

olya-2409 [2.1K]

Answer:

New Momentum of Ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

<u>Explanation:</u>

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Momentum of Ball A=2.2 $\frac{k g m}{s}$

Momentum of Ball B=4.4 $\frac{k g m}{s}$

Momentum of Ball C=11 $\frac{k g m}{s}$

Momentum of Ball D=15 $\frac{k g m}{s}$

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Initial Momentum of Ball B=4.4 $\frac{k g m}{s}$

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

=3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,

Change in momentum of Ball B $\Delta p=m \times \Delta v$

Where $\Delta p$ =change in momentum

m=mass of the ball B

$\Delta v$ =change in velocity ball B

And $\Delta v=v,$ since all balls, have same velocity

Thus the above equation, changes to

$\Delta p=m \times v$

Substitute all the values in the above equation we get

$\Delta p=6 \times 2.2$

$=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Result:

Thus the New Momentum of ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

3 0

3 years ago

Read 2 more answers

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

$\mu$ = 0.1

From the question; we can arrive at two cases;

That :

$m_{2} a _ \ {net} }= m_2g - T$ ----- equation (1)

$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a = $[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]$ g

80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0

3 years ago

an object is thrown into the air with an initial velocity of 12 meters per seconds from a platform that is raised 4 meters above

butalik [34]

Answer:

1.52 seconds

Explanation:

Step 1: identity the given parameters

Initial velocity (u) = 12m/s

Height above ground (h1) = 4m

Final velocity (V) = 0

Step 2: calculate the height travelled by the object from 4m height (h2).

V^2 = U^2 -2gh

0= 12^2-2(9.8*h)

2(9.8*h) = 12^2

19.6*h = 144

h = 144/19.6

h = 7.347 m

Total height above ground (ht) = 4m +7.347m = 11.347m

Step 3: calculate the time reach ground

T = √(2h/g)

T = √(2*11.347/9.8)

T= √(22.694/9.8)

T= √2.316

T= 1.52 seconds

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3 years ago

What is the law of variation of the period T of a simple pendulum

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The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum and $g=9.81 m/s^2$ is the gravitational acceleration. As we can see, the period of a simple pendulum depends only on its length.

3 0

3 years ago

A 78−kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 21° above the horizontal direction

Feliz [49]

Answer:

274N 0.41

Explanation:

As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.

then

<em>friction=mass x gravity x sin(21)</em>

Fr=78kg x 9.8m/s2 x sin(21)=274N

<em>friction= coefficient of kinetic friction x normal force of from the slope</em>

Fr= u x 78kg x 9.8m/s2 x cos(21)=274N

Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41

8 0

3 years ago