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valentinak56 [21]
3 years ago
6

HEY CAN ANYONE ANSWER DIS RQ!!!!!!

Physics
2 answers:
emmasim [6.3K]3 years ago
7 0

Answer:

Density is the mass to volume ratio of an object.

> It tells you how compact the mass is.

> Density = mass/volume

The density of water is 1 g/ml or 1 kg/L or 1000 kg/m3

• If an object has a density less than that of water, it

will float.

• If an object has a density more than that of water, it

will sink

Explanation:

hope this helps some

Ksivusya [100]3 years ago
5 0
The answer is:
If the object has a higher density that it will sink.
If it has a lower density than it will float.
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seropon [69]
The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical
8 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
3 years ago
What displacement in cm would occur with a 75 N/m spring if you placed a 300 N weight on the spring?
ollegr [7]

Surface tension=75N/m

Weight=300N

\\ \bull\tt\longmapsto Surface\:Tension=\dfrac{Weight}{Displacement}

\\ \bull\tt\longmapsto Displacement=\dfrac{Weight}{Surface\:Tension}

\\ \bull\tt\longmapsto Displacement=\dfrac{300}{75}

\\ \bull\tt\longmapsto Displacement=4m

\\ \bull\tt\longmapsto Displacement=400cm

6 0
2 years ago
A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,
AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

Q_{water}=Q_{methanol}

where:

Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})

Now replacing:

0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]

4 0
2 years ago
A block oscillating on a spring has period T = 2.8 s . (Note: You do not know values for either m or k. Do not assume any partic
olasank [31]

Answer:

Part a)

T = 3.96 s

Part b)

T = 1.98 s

Part c)

T = 2.8 s

Explanation:

As we know that time period of spring block system is given as

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

T = 2.8 s

Part a)

If the mass of the block attached is doubled

then we will have

T' = 2\pi\sqrt{\frac{2m}{k}}

T' = \sqrt2 T

T' = 3.96 s

Part b)

If the spring constant is doubled

then we have

T' = 2\pi\sqrt{\frac{m}{2k}}

T' = \frac{T}{\sqrt2}

T' = 1.98 s

Part c)

If the amplitude is halved but mass and spring constant will remain the same

so here we know that time period does not depends on Amplitude

so we will have

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

T = 2.8 s

7 0
2 years ago
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