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3 years ago

The time for a sound wave to travel between two people is 0.80 s,

Physics

1 answer:

sasho [114]3 years ago

4 0

Answer:

Explanation:

Using the below formula

Speed of sound = ( distance between observers) *2/(total time taken)

Now putt the given values ,

time taken = 0.80 sec

distance = 256 m

hence

V of sound= 256*2/0.80

V of sound = 640 m/sec

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A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel

spin [16.1K]

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

$m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s$

So Option d is correct one

5 0

3 years ago

A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

6 0

3 years ago

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th

Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

8 0

3 years ago

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0

3 years ago

Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro

Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0

3 years ago