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4 years ago

Plz can somebody explain for me how we do this exercice...thx

Physics

1 answer:

kompoz [17]4 years ago

7 0

1). Calculate how long it will take him to fall 5m to the ground.

2). THAT'S how long he has to cover the 24m horizontally, so you can calculate what his horizontal speed has to be.

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A rock is dropped in a pond, causing circular ripples. The radius increased with a rate 1 foot per second. When the radius is 4

Brrunno [24]

Answer:

true

Explanation:

Area = πr²

Differentiating in respect to r

we get,

dA/dr = 2πr

using chain rule,

$\frac{dA}{dt} = \frac{dA}{dr} .\frac{dr}{dt} = 2\pi \frac{dr}{dt}$

since r = 4feet

so ,

2π(4) = 8π

5 0

3 years ago

A cell phone sending a text message what kinda energy is to the scenario?

creativ13 [48]

Answer:

low powered radio frequency (RF) energy

7 0

3 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

4 years ago

The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and

ira [324]

Complete question is;

a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?

b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be

Answer:

A) V_t = 18 m/s

B) V_t = 10.39 m/s

Explanation:

Formula for terminal speed is given by;

V_t = √(2mg/(DρA))

Where;

m is mass

g is acceleration due to gravity

D is drag coefficient

ρ is density

A is Area of object

A) Now, for sphere 1,we have;

m = 1 kg

V_t = 6 m/s

g = 9.81 m/s²

Now, making D the subject, we have;

D = 2mg/((V_t)²ρA))

D = (2 × 1 × 9.81)/(6² × ρA)

D = 0.545/(ρA)

For sphere 2, we have mass = 9 kg

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]

V_t = 18 m/s

B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.

Thus;

Area of sphere 2 = 3A

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]

V_t = 10.39 m/s

5 0

4 years ago

a car traveling at 7.0 m/s accelerates uni formally at 2.5 m/s^2 to reach a speed of 12.0 m/s how long does it take for the acce

Ronch [10]

The problem wants to compute how long does it take for the acceleration to occur if a car travelling at 7m/s accelerates uniformally at 2.5 m/s^2 to reach a speed of 12 m/s and base on that, the time duration would be 2 seconds. I hope this would help

3 0

3 years ago