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2 years ago

The van accelerates at 2.2m/s^2. The force causing the acceleration is 5.5 kN. Calculate the mass of the van

1 answer:

8 0

The mass of the van will be 2.5 kg, Mass is found as the ratio of the force and the acceleration.

<h3>What is force?</h3>

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

F=ma

5.5 kN = m × 2.2m/s²

m= 2.5 kg

Hence,the mass of the van will be 2.5 kg,

To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

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What is the definition of energy

natita [175]

Energy (in Physics) is the ability to do work.

5 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

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

$\frac{v_o}{r_o} = \frac{v_1}{r_1}$

v1 = $\frac{r_1}{r_o} \ \ v_o$

let's calculate

v₁ = $\frac{1.50}{1.00} \ \ 25.0$

v₁ = 37.5 cm / s

4 0

3 years ago

Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its

sashaice [31]

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

$\omega=\sqrt{\dfrac{1}{LC}}$

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

6 0

3 years ago

A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

$V=1.4*10^5m/s$