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2 years ago

5

A 10 kg ball moving at 13 m/s strikes a 20 kg ball at rest. after the collision the 10 kg ball is moving with a velocity of 7m/s

.what is the velocity of the second ball?

2 answers:

Lostsunrise [7]2 years ago

6 0

Answer:

30 (kg)

Explanation:

therefore the mass of the ball is 2 so 30 (kg)

PolarNik [594]2 years ago

5 0

Answer:

3 m/s

Explanation:

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17. The edges of a rectangular solid have these measures: 1.5 feet by 1½ feet by 3 inches. What is its volume in cubic inches? a

Alisiya [41]

Answer:

c. 972

Explanation:

The volume of a rectangular solid is calculated as the product of its dimensions, that is, its width, its length and its height:

$V=a*b*h$

1 feet is equal to 12 inches, so:

$a=1.5ft*\frac{12in}{1ft}=18in\\b=(1+\frac{1}{2})ft\\b=\frac{3}{2}ft*\frac{12in}{1ft}=18in$

Now, we calculate the volume of the object in cubic inches:

$V=18in*18in*3in\\V=972in^3$

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2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

2 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

7 0

2 years ago

Read 2 more answers

A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

b)

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0

2 years ago

At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

vlabodo [156]

Answer:

$20.3125$ kJ/mol

Explanation:

$P_{i}$ = initial vapor pressure = 45.77 mm Hg

$P_{f}$ = final vapor pressure = 193.1 mm Hg

$T_{i}$ = initial temperature = 213.1 K

$T_{f}$ = final temperature = 243.7 K

$H$ = Heat of vaporization

Using the equation

$ln\left ( \frac{P_{f}}{P_{i}} \right ) = \left ( \frac{-H}{R} \right )\left ( \frac{1}{T_{f}} - \frac{1}{T_{i}}\right)$

$ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)$

$H = 20312.5$ J/mol

$H = 20.3125$ kJ/mol

8 0

2 years ago