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3 years ago

11

What is the momentum of a 120 pound bicyclist that is traveling at 25 mph?

1 answer:

klio [65]3 years ago

3 0

Momentum’s equation is p=mv

p=(120)(25)
p=3000

The biker’s momentum is 3000kgm/s

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Audible wavelengths. the range of audible frequencies is from about 20.0 hz to 2.00×104 hz . what is range of the wavelengths of

BabaBlast [244]

Given:
f1 = 20 Hz
f2 = 20000 Hz
speed of sound at 20 degrees celcius = 343 m/s

Solution:

for f1 = 20 Hz,

Using the equation:

lambda = speed of sound / f1 = 343 / 20 = 17.15 m

For f2:

lambda = speed of sound / f2 = 343 / 20000 = 0.01775 m

Therefore the wavelength range of audible sound in air would be 17.15 m to 0.01775 m.

6 0

3 years ago

Read 2 more answers

Proxima Centauri and the Sun are both main-sequence stars, but Proxima Centauri is a red dwarf and the Sun is a yellow dwarf. Ba

shutvik [7]

Answer:

Yellow dwarfs are small, main sequence stars. The Sun is a yellow dwarf. A red dwarf is a small, cool, very faint, main sequence star whose surface temperature is under about 4,000 K. Red dwarfs are the most common type of star. Proxima Centauri is a red dwarf.

Explanation:

6 0

3 years ago

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Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.

Sliva [168]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $|v| = 6.93$

Explanation:

From the question we are told that

The initial point is $(x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )$

The terminal point is $(x_2 , y_2 , z_2) = (-5 , 11, 8 )$

Generally the magnitude of the vector is mathematically represented as

$|v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}$

=> $|v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}$

=> $|v| = 6.93$

3 0

3 years ago

Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler

ehidna [41]

A) Acceleration in part A: $6.1 m/s^2$

B) Acceleration in part B: $2.7 m/s^2$

C) Acceleration in part C: $1.5 m/s^2$

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

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

where

v is the final velocity

u is the initial velocity

$\Delta t$ is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

$\Delta t = 0.9 - 0 = 0.9 s$

So the acceleration is

$a=\frac{5.5-0}{0.9}=6.1 m/s^2$

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

$\Delta t = 2.4 - 0.9 = 1.5 s$ is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

$a=\frac{9.5-5.5}{1.5}=2.7 m/s^2$

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

$\Delta t = 3.4 - 2.4 = 1 s$ is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

$a=\frac{11-9.5}{1}=1.5 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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3 0

3 years ago

A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 12 years have

tigry1 [53]

Answer:

$7.2878\times10^{16} m$

Explanation:

Time elapsed on Earth $d_{t}$ = 12 years

Time elapsed on board the ship $d_{t_o}$ = 9.2 years

now r= $\frac{d_{t} }{d_t_o}$ = 12/9.2= 1.3043

$v= c \sqrt{1-\frac{1}{r^2}}$

$v= 3\times10^8 \sqrt{1-\frac{1}{1.3043^2}}$

v= $1.9258\times10^8$

therefore distance L= $d_{t}\times v$

putting value we get

= $12\times\times1.9258\times10^8$

= $12\times\times365\times24\times3600\times1.9258\times10^8$

= $7.2878\times10^{16} m$

4 0

3 years ago