What is the object’s average acceleration between 0.0 s and 5.0 s? <br> ANSWER FAST! PLEASEE

A wave travels with speed 200 m/s. Its wave number is 1.5 rad/m. What are its (a) wavelength and (b) frequency

Vladimir [108]

Answer:

(a)

Explanation:

5 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago

A car with a velocity of 30 mph

Leokris [45]

Answer:

First calculate the deceleration making use of the formulation: V(very last speed= U(initial speed) -a(acceleration) t(ime) so 10=20- 4a. unfavourable a as decelerating. making a the acceleration the challenge by using rearranging the elements 4a= 20-10 = 10 so a=10/4 making use of Newton's 2d regulation P(stress) = M (mass) x A (acceleration). so P = 800 x 10/4 = 8000 /4 = 2000 newtons

Explanation:

4 0

3 years ago

The longest banana split ever made was 7.32 km long (obviously they used more than one banana). If an archer were to shoot an ar

elixir [45]

Answer:

The horizontal displacement of the arrow is not larger than the banana split.

Explanation:

Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.

So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.

y - y₀ = ut - 1/2gt²

0 - y₀ = 0 × t - 1/2gt²

-y₀ = -1/2gt²

t² = 2y₀/g

t = √(2y₀/g)

Substituting the values of the variables, we have

t = √(2y₀/g)

= √(2 × 8848.04 m/9.8 m/s²)

= √(17696.08 m/9.8 m/s²)

= √(1805.72 s²)

= 42.5 s

The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s

So, d = vt

= 100 m/s × 42.5 s

= 4250 m

= 4.25 km

Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.

8 0

3 years ago

What is a circuit breaker? What is a magnetosphere?

Umnica [9.8K]

A circuit breaker is an automatic device for stopping the flow of current in an electric circuit as a safety measure.

A magnetosphere is the region surrounding the earth or another astronomical body in which its magnetic field is the predominant effective magnetic field.

Hope this helps!! All credit goes to Google though

4 0

4 years ago

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What is the object’s average acceleration between 0.0 s and 5.0 s? ANSWER FAST! PLEASEE