PHYSICS COLLEGE

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air?
(b) What must have been the initial horizontal component of the velocity?
(c) What is the vertical component of the velocity just before the ball hits the ground?
(d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Answers

Answer 1

Answer:

Answer:

Explanation:

Given

height of building (h)=60 m

Range of ball=100 m

(a)time travel to cover a vertical distance of 60 m

$h=ut+(at^2)/(2)$

$60=0+(9.8* t^2)/(2)$

$t^2=12.24$

t=3.49 s

(b)To cover a range of 100 m

R=ut

$100=v_x* 3.49$

$v_x=28.57 m/s$

(c)vertical component of velocity just before it hits the ground

$v_y=u+at$

$v_y=0+9.81* 3.49=34.202 m/s$

(d) $v_(net)=√(v_y^2+v_x^2)$

$v_(net)=√(1986.02)=44.56 m/s$

Answer 2

Answer:

Final answer:

The ball is in the air for about 3.5 seconds. The initial horizontal velocity would have been approximately 28.6 m/s. The vertical component of the velocity just before the ball hits the ground is nearly 34.3 m/s. The overall velocity of the ball just prior to impact is roughly 44.6 m/s.

Explanation:

The problem given is about projectile motion which can be approached by splitting the motion into the horizontal and vertical components. We can work out the durations for each.

For the time the ball is in the air, we know that it falls vertically under gravity. Use the equation of motion, h = 0.5gt^2 where h is the height (60.0 m) and g is the acceleration due to gravity (approx 9.81 m/s^2). Solving for t takes approximately 3.5 seconds.
The initial horizontal component of the velocity can be calculated by the distance it traveled horizontally divided by the time it spent in the air. So, 100 m / 3.5 s = 28.6 m/s.
The vertical component of velocity just before the ball hits the ground can be calculated using v = gt where g is the acceleration due to gravity and t is the time. Solving gives around 34.3 m/s.
The overall velocity can then be calculated using the Pythagorean theorem, combining the horizontal and vertical components of the velocity - sqrt((28.6 m/s)^2 + (34.3 m/s)^2). This will approximately equal 44.6 m/s.

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А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?

Answers

Answer:

I am not able to answer this question please don't mind...

Explanation:

please marks me as brainliests...

Final answer:

The maximum expected measurement error for a pressure gauge measuring 0-10 bar with an inaccuracy of 1% of full-scale reading is 0.1 bar. When the gauge measures 1 bar, the expected inaccuracy is 10%.

Explanation:

The inaccuracy mentioned here is related to the full-scale reading which means the error is calculated based on the top measurement value. The pressure gauge range is 0-10 bar, so the inaccuracy is one percent of this. (a) Thus, the maximum measurement error expected for this instrument is 1.0% of 10 bar i.e., 0.1 bar. (b) If the gauge is measuring a pressure of 1 bar, then the relative error expressed as a percentage would be the absolute error (0.1 bar) divided by the observed reading (1 bar) i.e., 10%. It means, when measuring 1 bar pressure, the expected measurement error is 10%. This is an example of how instrument inaccuracy is properly interpreted and employed when working with various measurements.

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Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground afterSelect........seconds.

4.0

5.0

2.4

1.6

Answer and I will give you brainiliest

Answers

Answer:

2.4

Explanation:

Hope it help mark as Brainlist

If the diameter of the black marble is 3.0cm, and bye using the formula for volume, what is a good approximation if it’s volume? Record to the ones place

Answers

Complete question is;

If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?

Answer:

14 cm³

Explanation:

We will assume that this black marble has the shape of a sphere from online sources.

Now, volume of a sphere is given by;

V = (4/3)πr³

We are given diameter = 3 cm

We know that radius = diameter/2

Thus; radius = 3/2 = 1.5 cm

So, volume = (4/3)π(1.5)³

Volume ≈ 14.14 cm³

A good approximation of its volume = 14 cm³

A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answers

To solve this problem it is necessary to apply the concepts related to Newton's second law and the equations of motion description for acceleration.

From the perspective of acceleration we have to describe it as

$a = (\Delta v)/(\Delta t)$

Where,

$\Delta v$ = Velocity

$\Delta t$ = time

At the same time by the Newton's second law we have that

F = ma

Where,

m = mass

a = Acceleration

Replacing the value of acceleration we have

$F = m ((\Delta v)/(\Delta t))$

Our values are given as,

$m = 55*10^(-3)Kg$

$v = 46m/s$

$t = 0.00340s$

Replacing we have,

$F = m ((\Delta v)/(\Delta t))$

$F = (55*10^(-3))((46)/(0.00340))$

$F = 744.11N$

Therefore the magnitude of the average force exerted on the ball by the club is 744.11N

Explain what happent to the pressure exerted by an object when the area over which it is exerted:a) increase
b) decrease

Answers

The pressure decreases if the area increases. If the area decreases then the pressure increases.

2H is a loosely bound isotope of hydrogen, called deuterium or heavy hydrogen. It is stable but relatively rare — it form only 0.015% of natural hydrogen. Note that deuterium has Z = N, which should tend to make it more tightly bound, but both are odd numbers.Required:
Calculate BE/A, the binding energy per nucleon, for 2H in megaelecton volts per nucleon

Answers

Answer:

0.88 MeV/nucleon

Explanation:

The binding energy (B) per nucleon of deuterium can be calculated using the following equation:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u$

Where:

Z: is the number of protons = 1

N: is the number of neutrons = 1

$m_(p)$ : is the proton's mass = 1.00730 u

$m_(n)$ : is the neutron's mass = 1.00869 u

M: is the nucleu's mass = 2.01410

A = Z + N = 1 + 1 = 2

Now, the binding energy per nucleon for ²H is:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u = (1*1.00730 + 1*1.00869 - 2.01410)/(2)*931.49 MeV/u = 9.45 \cdot 10^(-4) u*931.49 MeV/u = 0.88 MeV/nucleon$

Therefore, the binding energy per nucleon for ²H is 0.88 MeV/nucleon.

I hope it helps you!

Final answer:

The binding energy per nucleon for 2H (deuterium) is 1.1125 MeV per nucleon.

Explanation:

The binding energy per nucleon, or BE/A, can be calculated by dividing the total binding energy of the nucleus by the number of nucleons. To calculate the BE/A for 2H (deuterium), we need to know the total binding energy and the number of nucleons in deuterium. The total binding energy of deuterium is approximately 2.225 MeV (megaelectron volts) and the number of nucleons is 2. Therefore, the BE/A for 2H is 2.225 MeV / 2 = 1.1125 MeV per nucleon.

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