A golf ball is dropped from rest from a height of 9.50 m. It hits the pavement, then bounces back up, rising just 9.70 m before falling back down again. A boy catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance calculate the total amount of time the ball is in the air, from drop to catch?

Answers

Answer 1

Answer: The ball is in the air for 27.70m because
9.5+9.7=19.2...
9.7-1.2=8.5...
19.2+8.5=27.70...

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Answer plz answer plzzz I am a little confused with full time

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I can’t read that I’m sorry make it more clear

An electron traveling horizontally to the right enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the electric field?

Answers

Answer:

Upward

Explanation:

For charged particles immersed in an electric field:

- if the particle is positively charged, the direction of the force is the same as the direction of the electric field

- if the particle is negatively charged, the direction of the force is opposite to the direction of the electric field

In this problem, we have an electron - so a negatively charged particle - so the direction of the force is opposite to that of the electric field.

Since the electric field is directed downward, therefore, the electric force on the electron will be upward.

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound

Answers

Answer:

about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

θ = arcsin(0.261538/1.03) ≈ 14.7°

At an angle of about 14.7°, someone outside the room will hear no sound.

How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

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Answer: hello your question is incomplete below is the complete question

Water stands at a depth H in a large open tank whose side walls are vertical . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

answer :

At Height ( h ) from the bottom of Tank

Explanation:

Determine how far above the bottom of the tank a second hole be cut

For the second hole to have the same range as the first hole

Range of first hole = Velocity of efflux of water * time of fall of water

= √ (2gh) * √( 2g (H - h) / g)

= √ ( 4(H-h) h)

Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank

Final answer:

The second hole should be cut at the same height as the first hole to have the same range for the stream.

Explanation:

In order for the stream emerging from the second hole to have the same range as the first hole, the second hole should be cut at the same height as the first hole. This is because the range of the stream depends on the initial velocity and the vertical distance traveled. If the second hole is higher or lower than the first hole, the vertical distance traveled will be different and the range of the stream will be affected.

Learn more about Range of stream from a hole here:

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An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places

Answers

Answer:

$4.26*10^6N$

Explanation:

A charge within an electric field E experiences a force proportional to the field whose module is F = qE, whose direction is the same, if the charge is negative, it experiences a force in the opposite direction to the field and if the charge is positive, experience a force in the same direction of the field.

In our case we are interested in the magnitude of the force, therefore the sign of the charge has no relevance

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact

Answers

Answer:

$y_(max) = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,(m)/(s))^(2) + 2\cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m)}$

$v = 2.913\,(m)/(s)$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m) + (1)/(2)\cdot (3\,kg)\cdot (2.913\,(m)/(s) )^(2) = (3\,kg) \cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s) ^(2)$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = (1)/(2)\cdot (2000\,(N)/(m) )\cdot (\Delta s)^(2) - (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot \Delta s$

$1000\cdot (\Delta s)^(2)-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_(1) \approx 0.128\,m$

$\Delta s_(2) \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s)^(2) = (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot y_(max)$