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2 years ago

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A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

he deposit. (Round your answer to three decimal places.)

1 answer:

vazorg [7]2 years ago

3 0

Answer:

8z

Explanation:

It is 8z

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A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag

klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

$\Phi =E\times S\times COS\Theta$

where Ф= flux

E= electric field

S= surface area

θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 $\frac{Nm^{2}}{sec}$

E= $1.44\times 10^{4}\frac{Nm}{C}$

S= $\pi \times 0.057^{2}$

$COS\Theta =\frac{\Phi }{S\times E}$

= $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

=0.5306

θ=57.94°

4 0

2 years ago

What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is $W=mg=2.28 N$

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

$F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N$

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

$F_y=W=mg=2.28 N$ (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

7 0

2 years ago

Read 2 more answers

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°

Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_{1} = 2.00\ L$

Pressure $P_{1}= 1.00\ atm$

Pressure $P_{2}= 70.0\ kPa$

Temperature $T_{1}= 20.0°C = 293\ K$

Temperature $T_{2}= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$\dfrac{PV}{T}=\ Constant$

For both temperature,

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}$

$V_{2}=\dfrac{101.325\times2\times280}{293\times70}$

$V_{2}=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

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2 years ago

A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters

JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0

2 years ago

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The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a freq

White raven [17]

Answer:92

Explanation:

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2 years ago