PLEASE HELP ME ASAP! IS MY ANSWER CORRECT?

A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is tu

kompoz [17]

Answer:

a

$F = 0.0566 \ N$

b

$t = 6.147 \ s$

Explanation:

From the question we are told that

The distance travel in 4.22 s is $s = 1.44 \ m$

The mass of the cart plus the fan is $m = 350 \ g = 0.35 \ kg$

Generally from kinematic equation we have that

$s = ut + \frac{1}{2} * a * t^2$

Here u is the initial velocity with value $u = 0 \ m/s$

So

$1.44= 0 * t + \frac{1}{2} * a * 4.22^2$

=> $a = 0.1617 \ m/s^2$

Generally the net force is

$F = m * a$

=> $F = 0.35 * 0.1617$

=> $F = 0.0566 \ N$

Gnerally the new mass of the cart plus the fan is $M = 656 \ g = 0.656 \ kg$

The distance considered is $s_1 = 1.63 \ m$

Generally the new acceleration of the cart is mathematically represented as

$F = M * a_1$

=> $a_1 = \frac{F}{M}$

=> $a_1 = \frac{0.0566}{0.656}$

=> $a_1 = 0.08628 \ m/s^2$

Gnerally from kinematic equation we have

$s = ut + \frac{1}{2} * a_1 * t ^2$

Here u is the initial velocity and the value is zero because it started from rest

=> $1.63 = 0 * t + \frac{1}{2} * 0.08628* t ^2$

=> $t = 6.147 \ s$

6 0

3 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

4 years ago

Someone please help<br> Just need answers

marishachu [46]

Answer:

As follows,

Explanation:

KE=1/2mv^2

In 1st question,

KE=1/2mv^2=1/2*0.05*12=0.3 J [50g=0.05 kg]

In 2nd question,

KE=1/2mv^2

6.8=1/2*0.046*v^2

v=sqrt(6.8/0.023)

v=17.19

In 3rd question,

KE=1/2mv^2

63/392=m

m=0.16kg=160g

For 4th,

a.

1st case,

KE=1/2mv^2=1/2*28*2.4^2=80.64

2nd case,

KE=1/2mv^2=1/2*28*3.7^2=191.66

Change in KE=191.66-80.64=11.02

b.Speed/velocity gained

6 0

3 years ago

A 9.0-cm-long spring is attached to the ceiling. when a 1.8 kg mass is hung from it, the spring stretches to a length of 18 cm .

Alex Ar [27]

The spring constant is computed by:
F = kx

Where: F is the force applied in newtons (N)

k is the spring constant measured in newtons per meter (N/m); and

x is the distance the spring is stretched (m)
and

F = mg

Where: F is the force pulling objects in the direction of the Earth.

m is the mass of the object.

g is the acceleration due to gravity;
So plugging our values in the formula:

F = mg

= (1.8) (9.81) = 17.658N

k =

F/x = 17.658 /0.09 = 196.2 N/meter

5 0

4 years ago

As water rises through the troposphere and cools, it changes from a(n) ————- to a(n) ———— through the process of condensation; w

Hoochie [10]

Answer:

GAS, LIQUID, CLOUDS

Explanation:

Water can be in three states: solid, liquid and gas, passing from one to another depending on the pressure and temperature.

In this complementation exercise.

When the water cools down the GAS goes to the LIQUID state, these small drops unite and form the CLOUDS

4 0

3 years ago