4: Question 2 - 10.0 pts possible

What would happen to mass and accelaration if the force on an object increases?? please help.

kozerog [31]

Wouldn't mass stay the same and acceleration increase or am I mistaken?

6 0

3 years ago

What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar

Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>

8 0

1 year ago

Erica (38 kg ) and danny (43 kg ) are bouncing on a trampoline. just as erica reaches the high point of her bounce, danny is mov

AfilCa [17]

<span>2.5 m/s going upward.
In the situation described, Erica and Danny undergo a non-elastic collision which will conserve their combined momentum. Since Erica is stationary, her momentum is 0. And since Danny is moving upward at 4.7 m/s his momentum is 43 kg * 4.7 m/s = 202.1 kg*m/s. Assuming that both Erica and Danny will be moving as a joined system, their combined mass is 38 kg + 43 kg = 81 kg. Since the momentum will be the same, their velocity will be 202.1 kg*m/s / 81 kg = 2.495061728 m/s. Since we only have 2 significant figures in the provided data, rounding the result to 2 significant figures gives a velocity of 2.5 m/s going upward.</span>

4 0

3 years ago

A person is riding a bicycle, and its wheels have an angular velocity of 10.7 rad/s. Then, the brakes are applied and the bike i

Scilla [17]

Answer:

(a) t = 22.9 s

(b) α= - 0.467 rad/s²

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated :

ωf ²= ω₀² + 2*α*θ Formula (1)

ωf= ω₀ + α*t Formula (2)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Data

θ = 19.5 revolutions : angular displacement of each wheel or angle that the wheel has rotated in a given time interval

ω₀= 10.7 rad/s : initial angular speed of the Wheel ( rad/s)

ωf = 0 : final angular speed of the Whee( rad/s)

Calculating of the angular acceleration (α )

We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :

ωf ²= ω₀² + 2*α*θ

(0 )²= (10.7)² + 2*α*(19.5*2*π )

0= 114.49 + (245.04)*α

-114.49 = (245.04)*α

α= (-114.49) /(245.04)

α= -114.49 /(245.04)

α= -0.467 rad/s²

Time does it take for the bike to come to rest

We replace data in the formula (2)

ωf = ω₀ + α*t

0 = 10.7 + -0.467*t

-10.7 = - 0.467*t we multiply by (-1) both sides of the equation :

10.7 = 0.467*t

t = 10.7 / 0.467

t = 22.9 s

3 0

3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

8 0

3 years ago