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3 years ago

Explain why ventilation is very important if there is risk of exposure to random gas in your home school

1 answer:

3 0

Ventilation is very important because it helps remove the gas form people’s homes and schools and it redirects the random gas outside so it is less likely to hurt people

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What is the formula to find PE and KE ?

Lapatulllka [165]

PE stands for Potential Energy. It is the stored energy in an object due to its position with respect to some reference. It is expressed in Joules.

P.E = m * g * h OR P.E. = mgh

m - mass of the body
g - acceleration due to gravity
h - height attained due to the body's displacement.

K.E. stands for Kinetic Energy. It is the energy possessed by a body due to its motion.

K.E = 1/2 mv² where m = mass of the body; v = velocity with which the body is moving.

5 0

3 years ago

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

muminat

Answer with Explanation:

We are given that

$y=(18-2x^2) km$

$x=(4t-3)km$

Differentiate x and y w.r.t t

$\frac{dx}{dt}=4$

$\frac{dy}{dt}=-4x\frac{dx}{dt}$

$\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)$

$v_x=\frac{dx}{dt}=4$

$v_y=\frac{dy}{dt}=-16(4t-3)$

Substitute t=1

$v_x=4$

$v_y=-16(4-3)=-16$

Magnitude of velocity= $\mid v\mid=\sqrt{v^2_x+v^2_y}$

$\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s$

Hence, the magnitude of the missile's velocity=16.49 m/s

$a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0$

$a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64$

Substitute t=1

$a_x=0,a_y=-64$

$\mid a\mid=\sqrt{a^2_x+a^2_y}$

$\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2$

Hence, the magnitude of acceleration when t=1 s= $64m/s^2$

3 0

3 years ago

How far does a car travel in 10 s if its initial velocity is 3 m/s and its acceleration

victus00 [196]

To learn more about Acceleration please visit -
brainly.com/question/13563385
#SPJ4

4 0

2 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

3 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

3 years ago