What services do plants provide? Select the three that apply.

A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes

Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0

3 years ago

compare to visible light, the wavelength of x-rays is shorter longer or the same and the frequency is lower higher or the same

Doss [256]

Answer:shorter higher

Explanation:Compare to visible light, the wavelength of X-rays is shorter and then frequency is higher

7 0

3 years ago

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

loris [4]

Answer:

The acceleration of the crate is $1.82\ m/s^2$ .

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

$F=ma\\\\F-f=ma$

f is frictional force, $f=\mu N=\mu mg$

$F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2$

So, the acceleration of the crate is $1.82\ m/s^2$ . Hence, this is the required solution.

4 0

2 years ago

Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot, isothermal metal substrate at Th = 300°C. The proper

Ad libitum [116K]

Answer:

74.52s

Explanation:

The solution is shown in the picture below

7 0

2 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

$\alpha=\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

b)

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

6 0

3 years ago