All categories

2 years ago

A bus of mass 3,500 kg that is moving at a speed of 18 m/s and

Physics

1 answer:

dezoksy [38]2 years ago

8 0

P₁ + P₂ = m₁V₁ + m₂V₂

3500(18) + 1300(20) =

63000 + 26000 =

890000 Kg • m/s

You might be interested in

In your own words describe how matter is defined

gizmo_the_mogwai [7]

Matter is any substance that has mass and takes up space.

5 0

3 years ago

A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

$V = \sqrt{\frac{T}{\mu}}$

Our values are given as:

$f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w$

Replacing we have that the velocity is

$V = \sqrt{\frac{T}{\mu}}$

$V = \sqrt{\frac{65}{0.230}}$

$V = 16.81m/s$

From the theory of wave propagation the average power wave is given as

$P =\frac{1}{2} \mu \omega^2 A^2 V$

Where,

A = Amplitude

$\omega = 2\pi f \rightarrow$ Angular velocity

$A^2 = \frac{2P}{\mu \omega^2 V}$

$A^2 = \frac{2P}{\mu (2\pi f)^2 V}$

Replacing,

$A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}$

$A = 0.0165m$

Therefore the amplitude of the wave should be 0.0165m

8 0

3 years ago

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu

galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

$F=I \times B \times L \times sin(\theta)$

So from data

$F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N$

Now sub parts

(a)

$\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N$

(b)

$\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N$

(c)

$\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N$

3 0

2 years ago

Help !!!! Estimate the number of breaths taken by a person during 44 years?

Lisa [10]

44 x 12. I got the 12 from the total of 12 months in a year.

44 > 40

x
12 > 10
----------
The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.

40 x 10 = 400.

Therefore, your estimate is 400.

The real answer is 520 breaths.

6 0

2 years ago

Help Please!!!!!!!!!!!!!!!!!!

julia-pushkina [17]

The answer is 59.4 degrees.

4 0

2 years ago