Ask question

Ask question

All categories

2 years ago

15

A 10 kg wagon is accelerated by a constant force of 60 N from an initial velocity of 5.0 m/s to a final velocity of 11 m/s. What

is the impulse received by the wagon? a)15 N s b)60 N s c)17 N s d)70 N s

1 answer:

Readme [11.4K]2 years ago

3 0

The answer is 274774

You might be interested in

Which wave has a greater frequency?

DedPeter [7]

The wave diagramed in blue.

8 0

2 years ago

Read 2 more answers

Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo

Vlad1618 [11]

Answer:

Tension T1 is less than tension T2.

T1 < T2

Explanation:

According to given data,

mass of box A ( mA) is grater than mass of box B (mB)

we can write,

m(A) > m(B)

Newton's second law states that:

Tension of object is directly proportional to the mass of the system.

T ∝ m

here Boxes A and B are being pulled to the right on a frictionless surface,

so Tension T1 generates due to the mass of box A m(A)

and Tension T2 arises due to mass of the system m(A) + m(B)

Thus tension T1 will be less than tension T2

T1 < T2

learn more about Tension force here:

<u>brainly.com/question/13175014</u>

<u />

#SPJ4

8 0

2 years ago

Which property of sound waves decreases as the square of the distance from the source increases?

AleksAgata [21]

A. Pitch

okay bye have a nice day

From: Charli

5 0

2 years ago

Read 2 more answers

Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an

galben [10]

Answer:

$\triangle P=1.95*10^{-4}$

Explanation:

Mass $m=0.001$

Diameter $d=1.2m$

Length $l=10m$

Generally the equation for Volume flow rate is mathematically given by

$Q=AV$

$V=\frac{Q}{\pi/4D^2}$

$V=\frac{0.001}{\pi/4(1.2)^2}$

$V=8.84*10^{-4}$

Generally the equation for Friction factor is mathematically given by

$F=\frac{64}{Re}$

Where Re

Re=Reynolds Number

$Re=\frac{pVD}{\mu}$

$Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}$

$Re=1040$

Therefore

$F=\frac{64}{Re}$

$F=\frac{64}{1040}$

$F=0.06$

Generally the equation for Friction factor is mathematically given by

$Head loss=\frac{fLv^2}{2dg}$

$H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}$

$H=19.9*10^{-9}$

Where

$H=\frac{\triangle P}{\rho g}$

$\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}$

$\triangle P=H*\rho g$

$\triangle P=1.95*10^{-4}$

6 0

3 years ago

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is $g = 167.2 \ m/s^2$

Explanation:

From the question we are told that

The length of the simple pendulum is $L = 1.081.08 \ m$

The number of cycles is $N = 101$

The time take is $t = 2.00 *10^{2 \ }s$

Generally the period of this oscillation is mathematically evaluated as

$T = \frac{N}{t }$

substituting values

$T = \frac{101}{2.0*10^2 }$

$T = 0.505 \ s$

The period of this oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{l}{g} }$

making g the subject of the formula we have

$g = \frac{L}{[\frac{T}{2 \pi } ]^2 }$

$g = \frac{4 \pi ^2 L }{T^2 }$

Substituting values

$g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }$

$g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }$

$g = 167.2 \ m/s^2$

7 0

3 years ago