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3 years ago

Which statement is true about two isotopes of the same element?

Physics

2 answers:

jarptica [38.1K]3 years ago

8 0

Answer:

D-They have different number of neutrons

Explanation:

brainliest? Plz

scZoUnD [109]3 years ago

3 0

Answer:

atoms that have the same # of protons but a DIFFERENT # of NEUTRONS

Explanation:

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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

Rahim is watching his favorite football team on television. In order to work, the television must be plugged into an electrical

Rasek [7]

I think the answer is B

8 0

3 years ago

Read 2 more answers

Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ

BartSMP [9]

Answer:

a) Δf = 0.7 n , e) f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

f = n / 2π (0.2335)² √ (210 10⁹/7800)

f = n /0.34257 √ (26.923 10⁶)

f = n /0.34257 5.1887 10³

f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no error is indicated in Young's modulus and density, so we will consider them exact

ΔE = Δρ = 0

Δf = df /dL ΔL

df = n / 2π √E /ρ | -2 / L³ | ΔL

df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

df = n 0.649

Absolute deviations must be given with a single significant figure

Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

f = n / 2π L² √ (E e /ρA)

where A is the area of the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

Δf = | df / dL | ΔL + df /de Δe + df /dA ΔA

Δf = 0.7 n + n 2π L² √(E/ρ A) | ½ 1/√e | Δe

+ n / 2π L² √(Ee /ρ) | 3/2 1√A23 |

the area is

A = b h

A = 24.9 3.3 10⁻⁶

A = 82.17 10⁻⁶ m²

DA = dA /db ΔB + dA /dh Δh

dA = h Δb + b Δh

dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

dA = (3.3 + 24.9) 0.005 10⁻⁶

dA = 1.4 10⁻⁷ m²

let's calculate each term

A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

A ’= n/ 2π L² √ (E /ρ) | ½ 1 / (√e/√ A) |Δe

A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

A '= 0.0266 n

A ’= 2.66 10⁻² n

A ’’ = n / 2π L² √ (E e /ρ) | 3/2 1 /√A³ |

A ’’ = n / 2π L² √(E /ρ) √ e | 3/2 1 /√ A³ | ΔA

A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

A ’’ = 6,738 10²

we write the equation of uncertainty

Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

Δf = 7 10² n

d) Δf = 7 10² n

e) the natural frequency n = 1

f = (15.1 ± 0.7) 10³ Hz

7 0

3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

3 years ago

Explain the relationship between mass volume and density

IrinaK [193]

Mass is a body of matter of indefinite shape and considerable size. Density is the degree to which something is filled. Check this website for more a better understanding : <span>www.answers.com/Q/What_is_the_relationship_between_density_and_volume</span>

8 0

3 years ago