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3 years ago

6

What is Speed? the ability to rest the ability to rest the ability to minimize the time cycle of a given movement the ability to

minimize the time cycle of a given movement the ability to apply force the ability to apply force the ability to use oxygen?

2 answers:

Vinvika [58]3 years ago

3 0

Answer:

Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time.

Explanation:

The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time.

natta225 [31]3 years ago

3 0

It seems you’ve already got an answer so yea

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A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0

sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

$v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})$

Since the initial height is 0

Solving for $v_{o}$

$v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s$

5 0

3 years ago

if you have an isotope of the element carbon (C) that has the mass number of 14, what other information would help you determine

n200080 [17]

The number of protons
number of neutrons=the mass number- number of protons
14-6=8

4 0

3 years ago

this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85

mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

$\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}$

(b)

$\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}$

(c)

$\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}$

3 0

10 months 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:

a.

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

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

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

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

A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy

snow_lady [41]

Answer:

<em> The distance required = 16.97 cm</em>

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

<em>e = 16.97 cm</em>

<em>Thus the distance required = 16.97 cm</em>

6 0

3 years ago