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4 years ago

This is a record of a marble being rolled along the ground. what was the marbles average speed for this entire trip brainly

Mathematics

2 answers:

yuradex [85]4 years ago

6 0

Answer:

The speed of marble is 1.5 m/s

Step-by-step explanation:

Consider the graph shown below.

As we know: $speed=\frac{Distance}{Time}$

To find the speed we need to substitute the value of distance and time by using the graph shown below.

From the graph we know when time was 0s marble covers a distance of 0 m.

When time was 2s marble covers a distance of 3 m.

Now substitute the value of distance and time in above formula.

$speed=\frac{3}{2}$

$speed=1.5$

Hence, the speed of marble is 1.5 m/s

saw5 [17]4 years ago

5 0

The answer is 1.5 m/s

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Answer:

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In 10 s, 200 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass

Dmitry_Shevchenko [17]

$\large\bf{\underline{Answer:}}$

$\large\bf{a) \triangle p_{1s} = - 120 \: kgm {s}^{ - 1} }$

$\large\bf{b) F = -120N}$

$\large\bf{c) Pressure=40.10\times 10^5 pa }$

__________________________________________

$\large\bf{\underline{In\: this\: problem\:we\:have:}}$

$\bf{N = 200\: bullets}$
$\bf{M= 5\times 10^{-3}kg}$
$\bf{V= 1200\:{ms}^{-1}}$

❒ To find the change in momentum for bullets , we need to remember the momentum p of a bullet is equal to product of mass and speed

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼p_{1}= mv}$

❒ This means , that change in momentum for one bullet will be equal to

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle P_{1} = mv_{f} - mv_{i}}$

$\large\bf{where\:v_{f}=0}$

Total change in momentum for the bullet in 10 sec is equal to product of change in momentum for one bullet and number of bullets hit the wall in 10 sec

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle P_{10s} = N\triangle P_{i}}$

Change in momentum given is the change of momentum in 10 sec is 10 times less

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p_{1s} = \frac{N\triangle p_{i}}{10}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p_{1s}=\frac{200.(mv_{f}-mv_{i}}{10}}$

$\large\bf{as\:said,v_{f}=0}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p_{1s}=\frac{-200.mv_{i}}{10}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p_{1s}=\frac{200.5\times 10^{-3}kg.1200ms^{-1}}{10}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p_{1s}=-1200\:Kgms^{-1}}$

__________________________________________

<h3>b) to find average force F on the wall we must remember that in general case force us the change of momentum in time :</h3>

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼F =\frac{\triangle P}{\triangle t}}$

Total change of momentum of bullets in 10 sec

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p =N\triangle p_{i} }$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p= N(mv_{f}-mv_{i})}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p= -N mv_{i}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p= -200.5\times 10^{-3}.1200ms^{-1}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼\triangle p= -1200kgms^{-1}}$

❒ We can find total force exerted in the wall in 10sec by dividing the momentum of bullet with 10 sec

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼F = \frac{\triangle p}{\triangle t}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼F = \frac{-1200}{10}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼F = -120N}$

__________________________________________

<h3>c) To find average pressure :</h3>

$\large\bf{area = 3\times 10^{-4}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼P=\frac{|F|}{A} }$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼P=\frac{-120}{3\times 10^{-4}}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟼P=40\times 10^4 Pa}$

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3 years ago

a disc is thrown into the air with an upward velocity of 62 ft/s.its height h in feet after t seconds is given by the function h

Korolek [52]

Given that the height of the disc is modeled by the function:
h=-16t^2+20t+6
a] The maximum height will be as follows:
At maximum height:
h'(t)=0
from h(t)
h'(t)=-32t+20=0
thus
t=20/32
t=5/8 sec
thus the maximum height will be:
h(5/8)=-16(5/8)^2+20(5/8)+6
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b] <span>How long will it take the disc to reach the maximum height?
</span>time taken to reach maximum height will be:
from h(t)
h'(t)=-32t+20=0
thus
t=20/32
t=5/8 sec
thus time taken to reach maximum height is t=5/8 sec

c]<span>How long does it take for the disc to be caught 3 feet off the ground?
</span>h(t)=-16t^2+20t+6
but
h(t)=3
thus
3=-16t^2+20t+6
solving for t we get:
0=-16t^2+20t+3
factoring the above we get:
t=5/8-/+√37/8
t=-1.5256 or 2.776
since there is no negative time we pick t=2.776
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