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2 years ago

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A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the ve

locity of the ball is 20.0 m/s. What is the magnitude of the
FINAL VELOCITY when it strikes the ground? (Note: Ignore air resistance.)

1 answer:

Charra [1.4K]2 years ago

5 0

Answer: 28.1

Explanation:

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A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0

2 years ago

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal

Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

p = m1 v1 + m2 v2

Part B.

Find the derivative

dp / dt = m1 dv1dt + m2 dv2 / dt

dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1 f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0

2 years ago

The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies

algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0

2 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅ = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0

2 years ago

If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel

Grace [21]

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

3 0

2 years ago