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3 years ago

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EX 6-1 A ball is twirled on a 0.870 - m-long string with a constant speed of 3.36 m / s . Calculate the acceleration of the ball

. Be sure to specify the direction of the acceleration.

1 answer:

Elina [12.6K]3 years ago

4 0

Answer:

$a=12.97\ m/s^2$

Explanation:

Given that,

The length of a string, l = 0.87 m

Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2$

So, the acceleration of the ball is $12.97\ m/s^2$ .

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What are the standard units of length, mass, and volume in the<br> metric system?

antoniya [11.8K]

Answer:

Meter, Gram and Liter.

Explanation:

In the metric system, the standard units for the below are;

Length - Meter

Mass - Gram

Volume - Liter.

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3 years ago

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s

Gekata [30.6K]

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

$P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580$

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2 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

aleksandr82 [10.1K]

Answer:

$A(t) = -340e^{-t/70} + 350$

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

$y = ce^{bt} + \frac{a}{b}$

So $A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350$

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

$A(t) = -340e^{-t/70} + 350$

4 0

3 years ago

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

Therefore the heat lost by mercury is 0.09J

5 0

3 years ago