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9 months ago

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A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum

1 answer:

Kryger [21]9 months ago

7 0

Given data

*The given mass of the pendulum is m = 3 kg

*The given height is h = 0.3 m

The formula for the maximum speed of the pendulum is given as

$v_{\max }=\sqrt[]{2gh}$

*Here g is the acceleration due to the gravity

Substitute the values in the above expression as

$\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}$

Hence, the maximum speed of the pendulum is 2.42 m/s

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A river flows due east with a speed of 3.00 m/s relative to earth. The river is 80.0 m wide. A woman starts at the southern bank

babymother [125]

Answer:

48 m

Explanation:

As she travels at the rate of 5m/s due north, the amount of time it would take for her to cross the 80m wide river would be

t = 80 / 5 = 16 seconds

This is also the time it takes for the river to push her to the east side at the rate of 3m/s. So after 16 seconds, she would reach the opposite point at a horizontal distance from her starting of

s = 16*3 = 48 m

4 0

3 years ago

A heated piece of metal cools according to the function c(x) = (.5)^(x _ 11), where x is measured in hours. A device is added th

Agata [3.3K]

You just need to replace x with 5 in each function

.5^5 - 11
-5-3

.5 ^-6
-8

64 - 8 = 56 A Celcius

Hope this helps

3 0

2 years ago

Read 2 more answers

A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

Alekssandra [29.7K]

Answer:

The electric field strength is $6.6\times10^{5}\ V/m$

Explanation:

Given that,

Magnetic field = 0.150 T

Speed $v= 4.40\times10^{6}\ m/s$

We need to calculate the electric field strength

Using formula of velocity

$v=\dfrac{E}{B}$

$E=v\times B$

Where, v = speed

B = magnetic field

Put the value into the formula

$E=4.40\times10^{6}\times0.150$

$E=660000\ V/m$

$E=6.6\times10^{5}\ V/m$

Hence, The electric field strength is $6.6\times10^{5}\ V/m$

4 0

2 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

A. A light wave moves through glass (n=1.5) at an angle of 25°. What angle will it have when it moves from the glass into air (n

torisob [31]

PART A)

By Snell's law we know that

$n_1sin i = n_2 sin r$

here we know that

$n_1 = 1.5$

$i = 25 degree$

$n_2 = 1$

now from above equation we have

$1.5 sin25 = 1 sin r$

$r = 39.3 degree$

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}$

$d_i = 30 cm$

so image will form at 30 cm from mirror

it is virtual image and smaller in size

3 0

3 years ago