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3 years ago

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A(n) 0.576 kg soccer ball approaches a player horizontally with a speed of 20.9 m/s. The player illegally strikes the ball with

her hand and causes it to move in the opposite direction with a speed of 9.73 m/s. What is the magnitude of the impulse delivered to the ball by the player?

1 answer:

Travka [436]3 years ago

4 0

Answer:

17.6kg.m/s in the opposite direction of the initial velocity.

Explanation:

In order to solve this we can use the Impulse-Momentum theorem:

$J=\Delta p\\J=m(v_f-v_o)$

We will assume the motion of the ball approaching the player as positive, so:

$J=0.576kg((-9.73m/s)-20.9m/s)=-17.6kg.m/s$

So the impulse is 17.6kg.m/s in the opposite direction of the initial velocity.

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A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H

IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm $f_{0} = 1210\ Hz$

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

$f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})$

Where, $f_{0}$ = frequency of source

$v_{0}$ = speed of observer

$v_{s}$ = speed of source

$v$ = speed of sound

Put the value into the formula

$f=1210\times\dfrac{343+(-8.25)}{343-0}$

here, source is at rest

$f=1180.8\ Hz$

$f=1181\ Hz$

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0

3 years ago

The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m

IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

$\rho$ = Resistivity of gold = $2.44\times 10^{-8}\ \Omega .m$ (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = $\dfrac{\pi}{4}d^2$

E = Electric field

Resistivity is given by

$\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m$

The electric field in the wire is 0.0360531138247 V/m

6 0

3 years ago

Read 2 more answers

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to

uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0

3 years ago

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

b.

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

3 years ago

There is a 90 kg woman attached at the end of a bungee cord (k = 35 N/m) that is experiencing simple harmonic motion. How long d

Cloud [144]

Answer:

The value is $T =10.1 \ s$

Explanation:

From the question we are told that

The mass of the woman is $m = 90 \ kg$

The spring constant of the bungee cord is $k = 35 \ N/ m$

Generally the period of the oscillation (i,e time taken to complete on cycle ) is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{m}{k} }$

=> $T = 2 * 3.142 * \sqrt{ \frac{90 }{ 35} }$

=> $T =10.1 \ s$

3 0

3 years ago