All categories

3 years ago

Plzz help me..plss I ll mark as a brainlist

Physics

1 answer:

Leni [432]3 years ago

5 0

Answer:

here are your answers friend....

Explanation:

Answer-1 hydrogen gas

Answer-2 egg cell or ovum

Answer-3 Estrogens

Answer-4 Gas Giants

Answer-5 carbon dioxide

Answer-6 Ganga action plan

Answer-7 Sputnik-1

Answer-8 Ductility

Answer-9 budding

Answer-10 liquefied petroleum gas

You might be interested in

What happened to the kinetic energy when the velocy of a moving body is doubled?

dem82 [27]

Answer:

If velocity is doubled, Kinetic Energy increases by 4 times. Kinetic energy of a body is the energy possessed by it, by virtue of its motion, i.e. if the body is moving it will always have kinetic energy.Nov 1, 2018

Explanation:

8 0

3 years ago

Read 2 more answers

Which activity would be the best choice foir a lifelong fitness program

Fantom [35]

I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.

7 0

3 years ago

Read 2 more answers

What happens to the density of ocean water when salinity increases? A. It increases. B. It decreases. C. It becomes variable. D.

omeli [17]

A. It increases... but slightly. Good luck!

3 0

3 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

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

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Plzz help me..plss I ll mark as a brainlist​

Plzz help me..plss I ll mark as a brainlist