On your business card , you position your businesses as "the area's most creative cookery " this phrase is your

A student conducts an experimenting to test how the temperature of a ball affects its bounce height. The same ball is used for e

uysha [10]

the independent variable is what you're testing or changing in an experiment, so the answer is the temperature of the ball when its dropped.

i hope that helped <3

3 0

2 years ago

Read 2 more answers

A cat is moving at 18 m/s when it accelerates for 2 seconds. What is its acceleration?

vova2212 [387]

Answer: acceleration: $4 m/s^{2}$

velocity: $26 m/s$

Explanation:

The complete question is written bellow:

A cat is moving at 18 m/s when it accelerates at $4 m/s^{2}$ for 2 seconds. What is his new velocity?

In this situation the following equation will be useful:

$V=V_{o}+at$

Where:

$V$ is the cat’s final velocity (new velocity)

$V_{o}=18 m/s$ is the cat’s initial velocity

$a=4 m/s^{2}$ is the cat's acceleration

$t=2 s$ is the time

Solving the equation:

$V=18 m/s+(4 m/s^{2})(2 s)$

$V=26 m/s$ This is the cat's new velocity

5 0

3 years ago

You are driving at 30.0 m/s on a freeway curve of radius 25.0 m. What is the magnitude of your acceleration?

mote1985 [20]

Answer:

36.0 m/s^2

Explanation:

4 0

3 years ago

From the given υ – t graph, it can be inferred that the object is

laila [671]

Answer:

a) in uniform motion

Explanation:

When velocity-time graph is parallel to x-axis, it shows that the object is moving with constant velocity. or we can say that that the object is moving with uniform motion. There is no change in velocity. It covere equal distance in equal intervals of time.

Hence, the correct option is (a) "in uniform motion".

6 0

2 years ago

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m

lesya [120]

Answer:

a) A =0.021525m

b) $\phi=0.37869rad$

c) $v_{max}=5.4098\frac{m}{s}$

d) $y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

Explanation:

1) Notation

A= Amplitude

v= velocity

$\lambda$ = wavelength

k= wave number

$\omega$ = angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

$y(t)=Acos(kx+ \omega t +\phi)$ (1)

At x=0 ,t=0 we have that:

$0.02=Acos(\phi)$

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

$v(t)=-\omega Asin(kx+ \omega t+\phi)$ (2)

And replacing the conditions at x=0, t=0 we got

$-2\frac{m}{s}=-\omega Asin(\phi)$

Now we can find the angular frequency with equation (3)

$\omega =\frac{2\pi}{T}$ (3)

Replacing the values obtained we got:

$\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}$

From equation (1) we have:

$Acos(\phi)=0.02$ (a)

$-2=-80\pi Asin(\phi)$ (b)

So from condition (b) we have:

$Asin(\phi)=\frac{1}{40\pi}$ (c)

If we divide condition (c) by condition (a) we got:

$\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789$

If we solve for $\phi$ we got:

$\phi =tan^{-1}(0.39789)=0.37869$

And now since we have $\phi$ we can find A from equation (a)

$Acos(0.37869)=0.02$

So then Solving for A we got $A=\frac{0.02}{cos(0.37869)}=0.021525$

3) Part c

From equation (2) we can see that the maximum speed occurs when $sin(\omega t+\phi)=1$ , so on this case we have:

$v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}$

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

$v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$ (4)

And solving for k from equation (4) we got

$k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}$

And with the k number we have everythin in order to create the wave function, given by:

$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

7 0

3 years ago