All categories

3 years ago

A bird flies at a speed of 4 m/s. It flies for 2160 seconds from its nest to the field. How much distance did the bird cover?

Physics

2 answers:

solniwko [45]3 years ago

8 0

Hello :D

For this applicate the formula:

$\boxed{d=vt}$

Data:

d = Distance = ?

v = Velocity = 4 m/s

t = Time = 2160 s

Replace according formula:

d = 4 m/s * 2160 s

Resolving:

d = 8640 m

Result:

The distance is of <u>8640 meters.</u>

Morgarella [4.7K]3 years ago

6 0

Answer:

The bird covers 8,640 m.

Explanation:

Distance= Speed×Time

Distance= 4×2160

Distance= 8,640

You might be interested in

How is the volume flow rate of water out of the tank, dVdt, related to the flow speed v ? Express your answer in terms of some,

Rainbow [258]

Answer:

$\frac{dV}{dt}= \frac{\pi d^2}{4}v$

Explanation:

The rate of volume flow out of tank can be expressed as:

$\frac{dV}{dt} = A\frac{dL}{dt}$

where,

dV/dt = Volume flow rate

A = Cross-sectional area of outlet = πd²/4

d = diameter of circular outlet

dL = Displacement covered by water

dt = time taken

but we know that:

Velocity = υ = displacement/time = dL/dt

Substituting the values of "dL/dt" and "A" in the equation, we get:

$\frac{dV}{dt} = \frac{\pi d^2}{4}v$

This is the expression for volume flow rate dV/dt, on terms pf v, d.

8 0

3 years ago

Do you wont to be friends

Angelina_Jolie [31]

Answer:

shore

Explanation:

8 0

3 years ago

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

3 years ago

in the design a thermos lab, you compared the temperature of your thermos with a container that you did not insulate. what was t

irina [24]

To see if the insulation would affect the temp of whatever you are measuring. <span />

4 0

3 years ago