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user100 [1]
2 years ago
10

A stone is thrown into a pond. Five waves were made from the same source in 10 seconds.

Physics
1 answer:
lina2011 [118]2 years ago
3 0

Answer:

The Frequency of the wave is 0.2Hertz

The period is 2seconds

Explanation:

We are asked to find the frequency and period

Frequency is the number of oscillations completed in one seconds

If five waves were made from the same source in 10 seconds, hence;

Frequency = number of oscillation/Time

Frequency = 5/10

Frequency = 0.5Hertz

Next is to get the period

Period = 1/Frequency

Period = 1/0.5

Period = 2seconds

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A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in o
mariarad [96]

Answer:

S_i=\frac{9}{10} =0.9

Explanation:

Given:

  • volume of oil in the cylinder, V_o=60\ cm^2
  • volume of the oil level when the ice is immersed, V=90\ cm^3
  • the volume level of oil when the ice melted, V'=87\ cm^3

<u>Now, therefore the volume of ice:</u>

V_i=V-V_o

V_i=90-60

V_i=30\ cm^3

<u>Now the volume of water:</u>

V_w=V'-V_o

V_w=87-60

V_w=27\ cm^3

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

S_i=\frac{\rho_i}{\rho_w} .....................(1)

as we know that density is given as:

\rm \rho=\frac{mass}{volume}

now eq. (1)

S_i=\frac{m}{V_{i}}\div  \frac{m}{V_w}

where, m = mass of the water or the ice which remains constant in any phase

S_i=\frac{V_w}{V_i}

S_i=\frac{27}{30}

S_i=\frac{9}{10} =0.9

7 0
2 years ago
The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N
kipiarov [429]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g  = 1.63 m/s^2

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

Weight=72 kg\times 1.63m/s^2=117.36 N

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

7 0
2 years ago
Read 2 more answers
A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0
Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

F_{lift}=F_{up} - F_{dw}=0.5*p*V^2

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

F_{lift}=0.5*p*V^2=W

By replacing the known values it is possible to find the plate's weight:

F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W

W=72.6 N

When the plate kept to 30° from the vertical the moment equation balance is written as:

F_{lift}=0.5*p*V^2=W*sen(30\°)

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

V=\sqrt(W*sen(30\°)/0.5p)

V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})

V=\sqrt(60.5 \frac{N}{kg/m^3})

V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})

V=\sqrt(60.5 \frac{m^2}{s^2})

V= 7.78 m/s

7 0
2 years ago
A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30
ElenaW [278]

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

7 0
2 years ago
Read 2 more answers
What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol
san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

Q=\frac{\Delta H}{n}

where,

Q = energy absorbed = 27.2 kJ

\Delta H = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

27.2kJ=\frac{6.01kJ/mol}{n}

n=0.221mol

Now we have to calculate the mass of ice.

\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}

Molar mass of ice = 18.02 g/mol

\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g

Thus, the mass of ice melted can be, 3.98 grams.

3 0
2 years ago
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