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Ostrovityanka [42]
3 years ago
9

How do you get better at physics

Physics
2 answers:
Greeley [361]3 years ago
4 0

Best way is to study, pay attention in class. Something i have found very helpful is take notes. Writing things down is a very good way to remember content.

Anit [1.1K]3 years ago
3 0
Study. Take notes as always. Highlight any thugs you don’t understand and I know ppl can be shy but ask your teachers for help. You might want to have extra help such as Some afterschool who provide extra help or even tutoring. Online videos are helpful too
You might be interested in
How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase
morpeh [17]

Answer:

The value is  v_o =  5.488 \  m/s

Explanation:

From the question we are told that

     The emitted frequency increased by \Delta f =  1.6 \% = 0.016 \

Let assume that the initial value of the emitted frequency is

      f =   100\%  =  1

Hence new frequency will be  f_n  =  1 +  0.016 = 1.016

Generally from Doppler shift equation we have that

         f_1 =  [\frac{ v \pm v_o}{v \pm + v_s } ] f

Here v  is the speed of sound with value  v =  343 \ m/s

         v_s is the velocity of the sound source which is v_s = 0 \ m/s because it started from rest

         v_o  is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

                   f_1 =  [\frac{ v + v_o}{v + 0 } ] f

=>                1.016  =  [\frac{ 343  + v_o}{343 } ] * 1  

=>                v_o =  5.488 \  m/s

         

8 0
3 years ago
Can someone help please and thank you:)
Ray Of Light [21]

Answer:

The answer is C.

Explanation:

4 0
3 years ago
A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa
scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

\text{Diameter of the lead sphere} =  D=1.940m

\text{Mass of the lead sphere} =  m_1 = 5681kg

\text{Mass of the dust particle} = m_2

Distance between the center of lead sphere to dust particle

r = \frac{D}{2}

r = \frac{1.940m}{2}

r = 0.97m

Gravitational force of the sphere on the dust particle:

F = \frac{Gm_1m_2}{r^2}

F = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5681kg)(m_2)}{(0.97m)^2}

F = (4.027*10^{-7} N/kg)m_2

Weight of the dust particle

W = m_2 g

W = m_2 (9.8m/s^2)

Ratio of F and W:

\frac{F}{W} = \frac{(4.07*10^{-7}N/kg) m_2)}{m_2(9.8m/s^2)}

\frac{F}{W} = 4.153*10^{-8}

Therefore the ratio is 4.153*10^{-8}

3 0
3 years ago
It’s your birthday, and to celebrate you’re going to make your first bungee jump. You stand on a bridge 100 m above a raging riv
amm1812

Answer:

X'=50.4\,m

Explanation:

Given that:

Height of jump, h=100\,m

length of elastic cord, l=30\,m

spring constant of the cord, k=40\,N.m^{-1}

mass of the body that jumps, m=80\,kg

Force on the bungee elastic cord:

F=m.g

F=80\times 9.8

Now this force F will be responsible for the elongation in the elastic cord, so:

F=k.x ............................(1)

where :

k = spring constant

x = extension in the elastic cord

using eq. (1)

80\times 9.8=40\times x

x=19.6\,m

So the cord stretches 19.6 meters more beyond its original length of 30 meters.

Hence, the remaining distance from the river surface at the bottom is:

X'=100-(30+19.6)

X'=50.4\,m

3 0
3 years ago
7) You think you have found a diamond. Its mass is 5.28 g and its volume is 2 cm3. Calculate the density
lesya [120]

Answer:

\boxed {\tt d=2.64 \ g/cm^3}

\boxed {\tt Not \ a \ diamond}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass is 5.28 grams and the volume is 2 cubic centimeters.

m=5.28 \ g\\v= 2 \ cm^3

Substitute the values into the formula.

d=\frac{5.28 \ g }{2 \ cm^3}

Divide.

d=2.64 \ g/c^3

The density of the unknown substance is 2.64 grams per cubic centimeter.

The density of a diamond is about 3.5 grams per cubic centimeter. Since 2.64 is not equal to 3.5, the unknown substance is not a diamond.

6 0
4 years ago
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