Which best illustrates projectile motion

Bird man is flying horizontally at a speed of 33m/s and a height of 86m. Bird man releases a turd directly above the start of th

Aliun [14]

Answer:

2838

Explanation:

8 0

3 years ago

How did Shi Huangdi link the new lands of his empire

Ierofanga [76]

Answer:

Many things, such as building the Great Wall of China, and building highways.

Explanation:

5 0

3 years ago

A copper wire 225m long must experience a voltage drop of less than 2.0v when a current of 3.5 a passes through it. compute the

MA_775_DIABLO [31]

Answer:

V = I * R

R = 2 / 3.5 = .571 ohms maximum resistance of wire

R = ρ L / A where R is proportional to L and inversely proportional to A

A = ρ L / R minimum area of wire

ρ = 1 / μ = 1.67E-8 ohm-m resistivity inverse of conductivity

A = 1.67E-8 ohm-m * 225 m / .571 ohm = 6.68E-6 m^2

A = 6.68 mm^2 since 1 mm^2 = 10-6 m^2 or 1 mm = 10-3 m

A = Π r^2 = 6.68 mm^2

r = (6.68 / 3.14)^1/2 mm = 2.13 mm radius of wire

d = 2 * r = 4.26 mm

5 0

2 years ago

A trailer truck with a 2000 [kg] cab and a 8000 [kg] trailer is traveling on a level road at 90 [km/hr].The brakes on the traile

NeX [460]

Answer:

a) t = 19.6 s, b) fr = 1.274 10⁴ N

Explanation:

This is a Newton's second law problem

Y Axis

for the cabin

N₁-W₁ = 0

N₁ = W₁

for the trailer

N₂- W₂ = 0

N₂ = W₂

X axis

for the cabin plus trailer, where friction is only in the cabin

fr = (m₁ + m₂) a

the friction force equation is

fr = μ N

we substitute

μ N₁ = (m₁ + m₂) a

μ m₁ g = (m₁ + m₂) a

a = μ g $\frac{m_1}{m_1 + m_2}$

let's calculate

a = 0.65 9.8 $\frac{2000}{2000+8000}$

a = 1,274 m / s²

a) to find the stopping distance we can use kinematics

Let's slow down the sI system

v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s

v = v₀ - a t

when it is stopped its speed is zero

0 = v₀ - at

t = v₀ / a

t = 25 / 1.274

t = 19.6 s

b) the friction force is

fr = 0.65 2000 9.8

fr = 1.274 10⁴ N

This is the braking force and also the forces that couple the cars.

3 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

4 years ago