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3 years ago

14

a 55 kg baseball player slides into third base with an initial speed of 4.6 m/s If the coefficient of kinetic friction between t

he player and the ground is 0.46, what is the player's acceleration?

1 answer:

Ronch [10]3 years ago

6 0

Hello there.

<span>0 + ½ m v(i)² - µ m g d = 0 + 0 </span>

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A body moving in the positive x direction passes the origin at time t = 0. Between t=0 and t=1 second, the body has a constant s

lord [1]

The important thing in this type of questions is to keep track of what you are doing at every point.

From t=0 to t=1s, there is uniform motion at v=24 m/s.

Therefore, it will move 24m in that second of time.

Afterwards, an acceleration comes in, so from t=1s to 11s there will be uniformly accelerated motion with acceleration -6m/s^2.

To account for this, you need to use Suvat's equation:

x(t)=x0+v0*t+1/2*a*t^2.

You should know what to plug in for each of the symbols in this equation:

x0 [initial position at t=1s] = 24m

v0 [initial velocity at t=1s] = 24 m/s

a [acceleration, switched on at t=1s] = -6 m/s^2

t [time from the start of the acceleration until the end, i.e. from t=1s to t=11s] = 10s

Plugging in those numbers in the equation will give you the position at t=11s.

4 0

3 years ago

Read 2 more answers

A blimp, suspended in the air at a height of 600 feet, lies directly over a line from a sports stadium to a planetarium. If th

MAXImum [283]

Answer:

1946 ft

Explanation:

The distance is the addition of the distance gotten from both triangles.

d = x + y

d = 1178 ft + 768 ft

d = 1946 ft

Attached is a picture showing how I arrived at the answer

3 0

3 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

3 years ago

When celery is placed in a glass of pure water the solution inside its cells is _____ compared to the water.?

kow [346]

The correct answer to the question above is hypertonic. When celery is being placed in a glass of pure water, the solution inside its cells is going to appear as hypertonic compared to the water. This means that the cells inside has a higher concentration than outside.

4 0

3 years ago

A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration o

Sergeeva-Olga [200]

Answer:

$a=2.4\ m/s^2$

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

$d=ut+\dfrac{1}{2}at^2$

or

$d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2$

So, the acceleration of the car is $2.4\ m/s^2$ .

5 0

3 years ago