Ask question

Ask question

All categories

3 years ago

5

A tree limb of mass 12 kg falls straight down. If air resistance exerts 27 N of force on the limb as it falls, what is the net f

orce on the tree limb

2 answers:

Mandarinka [93]3 years ago

8 0

Answer: 91N down

Explanation:

Karolina [17]3 years ago

6 0

The net force is 27 n

You might be interested in

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l

goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

$\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}$

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

$\Delta t = \frac{0.18\ m}{340\ m/s}$

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>

4 0

3 years ago

A ball rolls from x = - 5m to x = 0m in 1 second . What was its average velocity? (Units = m/s) Don't forget: velocities and dis

katrin2010 [14]

Answer:

+ 5 m/s

Explanation:

change in displacement = ΔX=final position - initial position

ΔX = 0-(-5) =0+5 =+ 5 m

average velocity = ΔX/t

= +5/1

= + 5 m/s

positive sign shows that ball rolls towards right

8 0

3 years ago

A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of

Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

1.29 kg/m³ = <u> mass </u>

1800 m³

mass = 1.29 kg/m³ ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0

3 years ago

Read 2 more answers

The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

Diano4ka-milaya [45]

Rx= 3.5 km

Ry= 2.9 km

4 0

3 years ago

Read 2 more answers

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago