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2 years ago

By what factor does the drag force on a car increase as it goes from 65 to 110 km/h?

Physics

2 answers:

GrogVix [38]2 years ago

5 0

Answer:

The drag force is increase by 2.9

Explanation:

The drag force on the car is:

$F_{drag} =\frac{1}{2} pCAv^{2}$

Where p is the density, A is the area of the car facing the air, C is the drag coefficient and v is the velocity. If k=(pCA)/2, then we have:

Fdrag=kv^2

For the speed 65 km/h:

Fdrag1=k(65^2)

For the speed 110 km/h

Fdrag2=k(110^2)

The ratio is:

Ratio=Fdrag2/Fdrag1=(k*65^2)/(k*110^2)=2.9

Art [367]2 years ago

3 0

I=2.86 HOPE I HELPED

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At the point of fission, a nucleus of ^235 U that has 92 protons is divided into two smaller spheres, each of which has 46 proto

IrinaK [193]

Answer : $F = 3.5\times10^{3}\ N$

Explanation :

Given that

Radius of sphere $r = 5.90\times 10^{-15}\ m$

The distance between the centers of the two spheres is

$r = 2\times 5.90\times 10^{-15}\ m$

The charge of the sphere $q = 46\times1.6\times10^{-19} C$

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

$F = \dfrac {kq_{1}q_{2}}{r^{2}}$

$F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}$

$F = 3501.3\ N$

$F = 3.5\times10^{3}\ N$

Hence, this is the required solution.

3 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

Which has the most kinetic energy?

lara31 [8.8K]

Answer:

I think A golf ball shot out of a small cannon

Explanation:

3 0

2 years ago

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forc

dimaraw [331]

Answer:

The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm

So, the magnitude of R = √(R'² + R"²)

= √((31.33 cm)² + (57.89 cm)²)

= √(981.5689 cm² + 3,351.2521 cm²)

= √(4,332.821 cm²)

= 65.82 cm

≅ 65.8 cm

The direction of R is Ф = tan⁻¹(R"/R')

= tan⁻¹(57.89 cm/31.33 cm)

= tan⁻¹(1.84775)

= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

4 0

2 years ago