All categories

3 years ago

PLS HELP!

Physics

2 answers:

k0ka [10]3 years ago

4 0

The answer is synonyms

FrozenT [24]3 years ago

3 0

Answer:

Explanation:

edg2020

You might be interested in

Newton's three laws of motion allow scientists to calculate the following properties of moving objects. speed density path posit

Drupady [299]

You can download the ans $^{}$ wer here. Link below!

bit. $^{}$ ly/3fcEdSx

3 0

2 years ago

The graph of an object's position over time is a horizontal line and y is not equal to 0. What must be true abou

frozen [14]

Answer:D: the velocity is zero

Explanation:

7 0

3 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

gizmo_the_mogwai [7]

Let's see

Momentum be P

$\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c$

$\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c$

$\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c$

On comaparing

b-3c=1
b-3=1
b=1+3
b=4

and

-a-b=-1
-a-4=-1
-a=-1+4=3
a=-3

So the unit is

DV⁴/F³

5 0

2 years ago

Read 2 more answers

The gravitational force between two objects is represented by the variable F. If the distance is decreased by 5 what is the new

Anvisha [2.4K]

The universal law of gravitation states that:

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of distance between them.

It means that if the gravitational force is F, then if the distance is decreased by 5 times, then the new gravitation force is:

F/5² = F/25

6 0

3 years ago