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Time measurements from a stopwatch are not precise. Why not ?

fgiga [73]

Human error (average human reaction time is .2 seconds)

5 0

3 years ago

A light-year is a unit of a. time. c. mass. b. distance. d. density. please select the best answer from the choices provided a b

11111nata11111 [884]

Answer:

distance

Explanation:

it is the distance traveled by light in one year

3 0

2 years ago

Read 2 more answers

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

3 0

3 years ago

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

Answer:

$2.78\times 10^{-35}\ \text{kg m/s}$

$6.178\times 10^{-34}\ \text{m/s}$

$0.31\times 10^{-4}\ \text{m/s}$

Explanation:

$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

h = Planck's constant = $6.626\times 10^{-34}\ \text{Js}$

m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

Golf ball minimum uncertainty in the momentum of the object

$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

Electron

$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $0.31\times 10^{-4}\ \text{m/s}$ .

6 0

3 years ago

1)what is the momentum of a 40,000,000kg tanker traveling at 5 m/s?

Fittoniya [83]

Answer:

1)4*10^7 * 5= 2*10^8

2)50/10=5

5 0

3 years ago