PHYSICS COLLEGE

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 1

Answer:

Answer:

2.4

Explanation:

Hope it help mark as Brainlist

Related Questions

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 516 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.6 ft deep, estimate the magnitude of his average acceleration as he slowed to a stop (that is while he was impacting the snow).
Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long range or short range.
In a solution such as salt water, the component that does the dissolving is called the?
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?A 0.31 sB 0.56 sC 4.3sD 70s
If you drew magnetic field lines for this bar magnet, which statement would be true

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 2.35 ✕ 105 A/m3. This function applies out to the wire's radius of 2.00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 11.5 μm and is at a radial distance of 1.20 mm?

Answers

The current contained within the width of a thin ring concentric is 18.1 x 10⁻⁶A

What is Current?

This is defined as electric charges moving through an electric conductor or space.

Parameters

Current density of J(r) = Br, where B = 2.35 x 10⁵ A/m³.

I = Jₓ A

where I is current, A is area and J is current density

A= 2rΔr

where 2r = circumference, Δr = width,

Substitute the values into the equation.

I=J(2rΔr)

I=2Br^2Δr

I= 2(2.35 x 10⁵)(1.2 x 10⁻³)^2(11.5 x 10⁻⁶)

= 18.1 x 10^-6 A

Read more about Current here brainly.com/question/25922783

Answer:

18.1 x 10^-6 A

Explanation:

A cylindrical wire carries a current density of J(r) = Br, where B = 2.35 x 10^5 A/m^3, to find the current within a certain area we multiply the current density with the are of this area:

I = J*A

for a ring with r distance from the center and width Δr, where Δr<< r, the area is:

A= 2 $\pi$ rΔr

where 2 $\pi$ r is the circumference and Δr is the width, substitute to get:

I=J(2 $\pi$ rΔr)

I=2 $\pi$ Br^2Δr

substitute with the given values to get:

I= 2 $\pi$ (2.35 x 10^5)(1.2 x 10^-3)^2(11.5 x 10^-6)

= 18.1 x 10^-6 A

A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her

Answers

Answer:

The average upward force exerted by the water is 988.2 N

Explanation:

Given;

mass of the diver, m = 60 kg

height of the board above the water, h = 10 m

time when her feet touched the water, t = 2.10 s

The final velocity of the diver, when she is under the influence of acceleration of free fall.

V² = U² + 2gh

where;

V is the final velocity

U is the initial velocity = 0

g is acceleration due gravity

h is the height of fall

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 10

V² = 196

V = √196

V = 14 m/s

Acceleration of the diver during 2.10 s before her feet touched the water.

14 m/s is her initial velocity at this sage,

her final velocity at this stage is zero (0)

V = U + at

0 = 14 + 2.1(a)

2.1a = -14

a = -14 / 2.1

a = -6.67 m/s²

The average upward force exerted by the water;

$F_(on\ diver) = mg - F_( \ water)\n\nma = mg - F_( \ water)\n\nF_( \ water) = mg - ma\n\nF_( \ water) = m(g-a)\n\nF_( \ water) = 60[9.8-(-6.67)]\n\nF_( \ water) = 60 (9.8+6.67)\n\nF_( \ water) = 60(16.47)\n\nF_( \ water) = 988.2 \ N$

Therefore, the average upward force exerted by the water is 988.2 N

A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m

Answers

Answer:

Explanation:

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

If 3.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.59 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t; ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5 = I x 2.59 x 3600

1469.5 = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

Clouds form from _____ temperature change, which occurs when an expanding gas cools.

Answers

The answer is Adiabatic

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.