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3 years ago

11

A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

ll and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone?

1 answer:

Hitman42 [59]3 years ago

3 0

Answer:

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2(0.15)v^2

1/2(0.15)v^2 = 70%*1/2(0.15)(20)^2

v^2 = 21/0.075

v^2 = 280

v = 16.73 m.s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m(v-u)

= 0.15(- 16.73-20)

= -5.51 kg*m.s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.

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Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan

Drupady [299]

Answer:

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

$0=0+C$

$C=0$

Substitute the value

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

4 0

3 years ago

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

$OB^2 = BC^2 + OC^2$

PUTTING ALL VALUE IN ABOVE EQUATION

$d^2 = 930-25sin30)^2 + (25cos30)^2$

$d^2 = 775$

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

3 0

2 years ago

In places of hot climate it is advised that the outer walls of houses be painted white. Explain.

goblinko [34]

Darker colors absorb heat, While lighter colors don't. If a house is painted black in Arizona, The black color will absorb the heat making the temperature inside the house very hot even with the AC on. If a house in Arizona is painted White, The heat will bounce off the White color, making the temperature inside the house cooler.

Hope this helps!

3 0

2 years ago

Read 2 more answers

A 45-mH inductor is connected in series with a 60-Ω resistor through a 15-V dc power supply and a switch. If the switch is close

egoroff_w [7]

Answer:

The current is 0.248 A

Explanation:

Given that,

Inductor $L= 45\times10^{3}\ H$

Resistance $R= 60\Omega$

Voltage = 15 volt

Time $t =7.0\times10^{-3}\ sec$

We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}(1-e^{\dfrac{-R}{L}}t)$

Where, V = voltage

R = resistance

L = inductance

T = time

Put the value into the formula

$I=\dfrac{15}{60}(1-e^{\dfrac{-60}{45\times10^{3}}}\times7\times10^{-3})$

$I=0.248\ A$

Hence, The current is 0.248 A.

4 0

3 years ago

2. A student drives 7.8-km trip to school and averages a speed of

Alekssandra [29.7K]

Answer:

<em>The total time is: t=451.22 sec</em>

<em>The average speed is: V=34.57 m/s</em>

Explanation:

<u>Average speed</u>

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

$\displaystyle V=\frac{x}{t}$

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

$\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec$

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

$\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec$

The total time is:

$t=239.26\ sec+211.96\ sec=451.22\ sec$

$\boxed{t=451.22\ sec}$

The average speed is:

$\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s$

$\boxed{\displaystyle V=34.57\ m/s}$

6 0

3 years ago