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2 years ago

A person walks 9 meters to the right and 8 meters to the left what's the distance and displacement

Physics

2 answers:

nordsb [41]2 years ago

5 0

The distance is 17 and the displacement is 1

saw5 [17]2 years ago

4 0

Answer:

walked: 17 meters

displacement: 1

Explanation:

so you add 9+8 and the displacement is 1 b/c when you moved back those 8 meters you were still 1 meter away from where you where originally standing

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An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the ma

Jet001 [13]

Answer:

Explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

because of negative charge on electron .

7 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas

Arada [10]

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

7 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.