Ask question

Ask question

All categories

3 years ago

9

If the horizontal component of a vector is 6 m/s and the vertical component is also 6 m/s, what is the resultant value of the ve

ctor?
A. 6 m/s
B. 8.5 m/s
C. 10.4 m/s
D. 12 m/s

1 answer:

BlackZzzverrR [31]3 years ago

6 0

Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.

You might be interested in

Which of the following is NOT a characteristic of an inner planet?. . A.. rocky. . B.. solid surface. . C.. near the sun. . D..

sveticcg [70]

Being made mostly of gas is NOT a characteristic of an inner planet. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

3 0

3 years ago

Read 2 more answers

Select the correct answer?

yKpoI14uk [10]

Answer:

The answer is d i think so yea

Explanation:

hope u hae a really good day

8 0

3 years ago

A grain silo is shown below:grain silo formed by cylinder with radius 5 feet and height 175 feet and a half sphere on the topwha

nekit [7.7K]

Find the volume of the bottom and top separately and then add them.
Cylinder volume is the area of the bottom times the height
(22/7)(5^2)•175=13750 ft^3

The volume of a sphere is
V=(4/3)(22/7)r^3
where r is the radius. Here that's also 5 since it fits on the cylinder.
Also we only want half the sphere so use
V=(2/3)(22/7)•5^3=261.9 ft^3
Which we round upto 262.
Now add the parts together
13750+262=14,012 ft^3

6 0

3 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 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