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4 years ago

Which statement BEST compares the force of gravity on a school bus on a school day and a weekend?

Physics

1 answer:

Lostsunrise [7]4 years ago

4 0

Answer:

The force of gravity between the Earth and the school bus is more on a school day than on a weekend

Explanation:

The force of gravity, F, is the attractive force that holds objects in place on the Earth's surface

The force of gravity attracting the bus of mass m to the Earth of mass M having a radius of $R_{Earth}$ , can be presented by the following equation;

$F = \dfrac{G \times M \times m}{(R_{Earth})^2}$

Which shows that the force increases as the mass of the bus increases

Since the school bus mass on a school day when carrying Pupils is mare than the mass of the bus in the weekend when the bus is empty, the force of gravity between the Earth and the school bus is more on a school day than on a weekend.

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An example of a decomposition reaction is

OlgaM077 [116]

A.is an example of decomposition reaction.

6 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

BigorU [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$f= -75 \ cm = - 0.75 \ m$

$P = -1.33 \ diopters$

Explanation:

From the question we are told that

The image distance is $d_i = -75 cm$

The value of the image is negative because it is on the same side with the corrective glasses

The object distance is $d_o = \infty$

The reason object distance is because the object father than it being picture by the eye

General focal length is mathematically represented as

$\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}$

substituting values

$\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}$

=> $f= -75 \ cm = - 0.75 \ m$

Generally the power of the corrective lens is mathematically represented as

$P = \frac{1}{f}$

substituting values

$P = \frac{1}{-0.75}$

$P = -1.33 \ diopters$

7 0

3 years ago

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat given to the system in thes

Mashcka [7]

Answer:

$\\ \\ \\ \\ \\ ....$

6 0

2 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

4 years ago