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3 years ago

You decide to take your middle school science class to the bowling alley for some “hands-on”

Physics

1 answer:

Reika [66]3 years ago

6 0

Answer:

(1) is incorrect and (2) is correct.

Explanation:

There is a force of friction opposing the motion, but as soon as Jill let go of the ball, there is no force pushing the ball onward. Therefore, the only forces acting on the ball is the force of friction and air resistance. (still friction)

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Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?

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Explain why it is not advisable to use small values of incident ray in performing experiment on therefraction through a glass prism.

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4 years ago

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You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude

Montano1993 [528]

The ball's acceleration is constant in magnitude and direction, from the instant it leaves your hand, until the instant it hits the ground, no matter what direction or speed you throw it.

It's the acceleration of gravity, on whatever planet you happen to be standing when you throw the ball.

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3 years ago

A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley c

sergiy2304 [10]

Answer:

Part a)

$a_{hanger} = g - \frac{T}{m}$

Part b)

$a_{cart} = \frac{T}{M}$

Part c)

$a = \frac{mg}{M + m}$

Part d)

$a = 1.35 m/s^2$

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

$mg - T = ma$

so we have

$a_{hanger} = g - \frac{T}{m}$

Part b)

now for car that is rolling on the floor the net force is given as

$F = Ma$

$T = Ma$

$a_{cart} = \frac{T}{M}$

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

$T = Ma$

$mg - Ma = ma$

$a = \frac{mg}{M + m}$

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

$a = \frac{0.50(9.81)}{2.40 + 0.50}$

$a = 1.35 m/s^2$

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3 years ago

"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1

kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Angular Speed formula as follows:

$\boxed{ \omega = \omega_o + \alpha t }$

$\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }$

$\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }$

$\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }$

where :

ω = final angular speed ( rad/s )

ω₀ = initial angular speed ( rad/s )

α = angular acceleration ( rad/s² )

t = elapsed time ( s )

θ = angular displacement ( rad )

$\texttt{ }$

Given:

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

Asked:

average angular speed = ?

Solution:

Firstly , we will calculate angular displacement as follows:

$\theta = \frac{( \omega + \omega_o )}{2} t$

$\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0$

$\theta = 10 \times 4.0$

$\boxed {\theta = 40 \texttt{ rad}}$

$\texttt{ }$

Next , we could calculate the average angular speed as follows:

$\texttt{average angular speed} = \theta \div t$

$\texttt{average angular speed} = 40 \div 4.0$

$\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Moment of Inertia : brainly.com/question/13796477
The Ratio of the Moments of Inertia : brainly.com/question/2176655

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0

3 years ago

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What traction of the radioisotope remains in the body after one day?

r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is $(\frac{1}{2})^{\frac{1}{\tau}}$ , with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

$m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where:

$m_0$ is the mass of the radioactive sample at t = 0

$\tau$ is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

$\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}$

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

$\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}$

where the half-life $\tau$ must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

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3 years ago