Ask question

Ask question

All categories

3 years ago

11

assuming birdman flies at height of 72m, how fast should he fly to hit bucket at 63m from start of field. gravity is -9.8m/s^2 n

o air resistance drops thing at start of field. what is his speed/velocity????????

1 answer:

algol [13]3 years ago

7 0

Answer:

.....Birdman?

Explanation:

You might be interested in

A spring has a force constant of 500.0 N/m. Show that the potential energy stored in the spring is as follows: a. 0.400 J when t

kogti [31]

Answer:

(a) Hence, the potential energy = 0.400 J.

(b) Hence the potential Energy = 0.225 J.

(c) Hence the potential energy = 0 J

Explanation:

Potential Energy; This is the energy of a body, by virtue of its position in the gravitational field. The unit of potential energy is Joules (J)

the potential energy stored in a spring is

Ep = 1/2ke²..................................... Equation 1

Where Ep = potential Energy, k = force constant, e = extension.

a.

When spring is stretched 4.00 cm, e = 4.00 cm = 4/100 = 0.04 m, and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0.04)²

Ep = 0.4 J.

Hence, the potential energy = 0.400 J.

(b)

When the spring is compressed 3.00 cm, e = 3.00 cm = 3/100 = 0.03 m, and k = 500 N/m.

Substitute into equation 1,

Ep = 1/2(500)(0.03)²

Ep = 250(0.0009)

Ep = 0.225 J.

Hence the potential Energy = 0.225 J.

(c) When the spring is unstretched, e = 0 cm = 0 m and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0)²

Ep = 250(0)

Ep = 0.

Hence the potential energy = 0 J

3 0

3 years ago

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

Ilya [14]

Answer with Explanation:

We are given that

Area of loop= $(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R= $0.1\Omega$

B= $4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf , $E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$

Current, $I=\frac{E}{R}$

Current, $I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid$

Substitute t=0 s

Then, I= $1.6\mid (1-0)\mid$ =1.6 A

Substitute t=1 s

Then, I= $1.6\mid (1-1)\mid$ =0

Substitute

t=2 s

Current, I= $1.6\mid(1-2)\mid$ =1.6 A

8 0

3 years ago

Read 2 more answers

A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled

Leya [2.2K]

Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer: 44.25 ft-lb

6 0

3 years ago

Two dogs are pulling on a chew toy. One dog pulls the chew toy with 64 N [E] and

PIT_PIT [208]

Answer:

Eastward, at 11 m/s^2

Explanation:

64N-31N=unbalanced force of 33N

F=ma

33N=(3kg)a

a=11m/s^2 to the East

3 0

2 years ago

Please awnser and show the ways

topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0

3 years ago