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2 years ago

PLEASE HELP!!!!

Physics

1 answer:

MrRissso [65]2 years ago

7 0

Answer:

6 cereal bars = 222g, which is much less than 1kg

Explanation:

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You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi

Monica [59]

Answer:

a) x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

2 = 10t + (2/3)*t^3

0 = 10t + (2/3)*t^3 + 2

- solve for t:

t = 0.1875 s

- Evaluate x* at t = 0.1875 s

x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

x*(0.1875) = 10.18 m

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3 years ago

An object has a mass of 60 kg. It decelerates from 50 m/s to 20 m/s when a resultant force of 300 N acts on it. For how long doe

Pavlova-9 [17]

Force=300N
Mass=60kg

Find Acceleration

$\\ \rm\longmapsto F=ma$

$\\ \rm\longmapsto a=\dfrac{F}{m}$

$\\ \rm\longmapsto a=\dfrac{-300}{60}$

$\\ \rm\longmapsto a=-5m/s^2$

Now

u=50m/s
v=20m/s

Using 1st equation of kinematics

$\\ \rm\longmapsto v=u+at$

$\\ \rm\longmapsto t=\dfrac{v-u}{a}$

$\\ \rm\longmapsto t=\dfrac{20-50}{-5}$

$\\ \rm\longmapsto t=\dfrac{-30}{-5}$

$\\ \rm\longmapsto t=6$

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C im pretty sure?.........

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Answer:

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Explanation:

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