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3 years ago

Anyone know how to do this

Physics

1 answer:

Charra [1.4K]3 years ago

8 0

The answer to your question is: Yes.

To solve each and every one, write the two given numbers in place of their letters in the given equation, and then solve the equation for the letter that's left.

For example, let's do #1 together. I'll do all the work, and you just watch.

#1 says that s=d/t. It also tells us that 'd' is 4 and 't' is 2. So here's a great idea: Why don't we write '4' where it says 'd', and write '2' where it says 't', and then see what we've got ?

s = 4/2

Wow ! We know that 4/2 is 2 . So s=2 and that's exactly the answer to the question !

Now it's YOUR turn. Do the next 25 problems the same way.

You might be interested in

How are electrical signals transmitted over long distances?

forsale [732]

Answer:

Over such small distances, digital data may be transmitted as direct, two-level electrical signals over simple copper conductors. This results from the electrical distortion of signals traveling through long conductors, and from noise added to the signal as it propagates through a transmission medium.

4 0

2 years ago

Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es

mojhsa [17]

Answer:

D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

D = 2 3960 1100 + 1100²

D = 8.712 10⁶ + 1.21 10⁶

D = 9.92 10⁶ mi

D = 9.9 10⁶ mi

8 0

3 years ago

How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to

Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.

3 0

2 years ago

Read 2 more answers

A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by

Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0

2 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago