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3 years ago

An ant can travel approximately 30 meters per minute. How many meters could an ant move in 45 minutes.

Physics

1 answer:

Nastasia [14]3 years ago

8 0

Answer:

From what I see, it's saying that every minute, the ant can move 30 meters. So how many meter would it move in 45 minutes?

30 meters = 1 min

x meters = 45 min

1 min x 45 = 45 min

30 meters x 45 = 1,350 meters

So, I believe the answer would be 1,350 meters.

hope this helps. :>

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The following table lists the work functions of a few commonmetals, measured in electron volts.

steposvetlana [31]

Answer:

Lithium

Explanation:

The equation for the photoelectric effect is

$\frac{hc}{\lambda}= \phi + K_{max}$

where

$\frac{hc}{\lambda}$ is the energy of the incident photon, with

h being the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

$\phi$ is the work function of the metal (the minimum energy needed to extract the photoelectron from the metal)

$K_{max}$ is the maximum kinetic energy of the emitted photoelectrons

In this problem, we have

$\lambda= 190 nm = 1.9\cdot 10^{-7}m$ is the wavelength of the incident photon

$K_{max}=4.0 eV$ is the maximum kinetic energy of the electrons

First of all we can find the energy of the incident photon

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.90\cdot 10^{-7} m}=1.05\cdot 10^{-18} J$

Converting into electronvolts,

$E=\frac{1.05\cdot 10^{-18} J}{1.6\cdot 10^{-19} J/eV}=6.6 eV$

So now we can re-arrange the equation of the photoelectric effect to find the work function of the metal

$\phi = E-K_{max}=6.6 eV - 4.0 eV=2.6 eV$

So the metal is most likely Lithium, which has a work function of 2.5 eV.

3 0

3 years ago

what is the average speed of (a) a car that travels 400m in 20s. and (b) an athlete who runs 1500m in 4 minutes

zmey [24]

Answer:

a) 20 m/s

b) 37.5 m)s

Explanation:

Average speed = total distance ÷ total time

=> (a) average speed of a car that travels 400m in 20s

= 400/20 = 20 m/s

& (b) average speed of an athlete who runs 1500m in 4 minutes (or 4×60=240 seconds)

= 1500/240 = 37.5 m/s

5 0

3 years ago

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

3 years ago

2. While standing near a bus stop, a student hears a distant horn beeping. The frequency emitted by the horn is 440 Hz. The bus

jarptica [38.1K]

Given Information:

Frequency of horn = f₀ = 440 Hz

Speed of sound = v = 330 m/s

Speed of bus = v₀ = 20 m/s

Answer:

Case 1. When the bus is crossing the student = 440 Hz

Case 2. When the bus is approaching the student = 414.9 Hz

Case 3. When the bus is moving away from the student = 468.4 Hz

Explanation:

There are 3 cases in this scenario:

Case 1. When the bus is crossing the student

Case 2. When the bus is approaching the student

Case 3. When the bus is moving away from the student

Let us explore each case:

Case 1. When the bus is crossing the student:

Student will hear the same frequency emitted by the horn that is 440 Hz.

f = 440 Hz

Case 2. When the bus is approaching the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330+20 )

f = 440 ( 330/ 350 )

f = 440 ( 0.943 )

f = 414.9 Hz

Case 3. When the bus is moving away from the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330-20 )

f = 440 ( 330/ 310 )

f = 440 ( 1.0645 )

f = 468.4 Hz

6 0

3 years ago

A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.

cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

$a=1.5 m/s^2$

So we can re-arrange the equation above to find their combined mass:

$m=\frac{F}{a}=\frac{75}{1.5}=50 kg$

3 0

3 years ago