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3 years ago

Can a body have an east velocity while expressing westward acceleration

Physics

1 answer:

Vlad [161]3 years ago

8 0

Yes it can, an object can be moving a certain direction while the ACCELERATION is in the opposite direction.

Lets say your riding a bike... if your squeezing your handle bar breaks, the acceleration of the bike would be pushing in the opposite direction of the direction the bike is moving.

Hope this helped!

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Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = 315.876 m/s

the root mean square velocity of the oxygen gas is

478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank

6 0

3 years ago

The energy emitted from the sun is a product of ________.

Tamiku [17]

Q. The energy emitted from the sun is a product of ________.

A. Fusion

3 0

4 years ago

The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons

bulgar [2K]

Answer: $\lambda =248.99 nm$

Explanation:

Given

Work function $\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98$

$h=6.626\times 10^{-34} J$

$c=2.998\times 10^8$

$\phi =\frac{hc}{\lambda }$

$\lambda =\frac{hc}{\phi }$

$\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}$

$\lambda =248.99 nm$

3 0

4 years ago

Which quantum number gives the approximate energy of an electron in the quantum mechanical model of the atom? A) n B) I C) m1 D

alexdok [17]

Answer:

Option A) n

Explanation:

In accordance to Quantum Mechanical model of an atom:

The Principle Quantum number (n) gives the description of the shell of an electron and the energy level of an electron in an atom.
The angular momentum also referred to as Azimuthal Quantum number (l) gives the description of the shape of the orbitals and helps in determination of angular momentum magnitude.
The magnetic quantum number ( $m_{1}$ ) describes the energy levels or the number of orbitals contained in a subshell and the way these are oriented within.

The spin quantum no. ( $m_{2}$ ) determines the elelctron spin's direction which may be ( $-\frac{1}{2}$ ) or ( $+\frac{1}{2}$ ).

4 0

3 years ago

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$