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1 year ago

11

a car with a mass of 2,00 kg travels a distance of 400 m as it moves from one stoplight to the next . as its fastest , the car t

ravels at 18 m/s. what is its kinetic energy at this point?

1 answer:

Semmy [17]1 year ago

7 0

Answer:

32400 J

Explanation:

K.E.=(1/2)(200)(18^2)

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A ball rolls from x = - 5m to x = 0m in 1 second . What was its average velocity? (Units = m/s) Don't forget: velocities and dis

katrin2010 [14]

Answer:

+ 5 m/s

Explanation:

change in displacement = ΔX=final position - initial position

ΔX = 0-(-5) =0+5 =+ 5 m

average velocity = ΔX/t

= +5/1

= + 5 m/s

positive sign shows that ball rolls towards right

8 0

3 years ago

A batter hits a fly ball which leaves the bat 0.89 m above the ground at an angle of 62 ∘ with an initial speed of 29 m/s headin

KatRina [158]

consider the motion in Y-direction

v₀ = initial velocity = 29 Sin62 = 25.6 m/s

a = acceleration = - 9.8 m/s²

t = time of travel

Y = vertical displacement = - 0.89 m

using the equation

Y = v₀ t + (0.5) a t²

- 0.89 = (25.6) t + (0.5) (- 9.8) t²

t = 5.3 sec

consider the motion along the horizontal direction :

v₀ = initial velocity = 29 Cos62 = 13.6 m/s

a = acceleration = 0 m/s²

t = time of travel = 5.3 sec

X = horizontal displacement =?

using the equation

X = v₀ t + (0.5) a t²

X = (13.6) (5.3) + (0.5) (0) t²

X = 72.1 m

d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m

t = time taken = 5.3 sec

v = speed of center fielder

using the equation

v = d/t

v = 34.9/5.3

v = 6.6 m/s

3 0

2 years ago

A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t

Ganezh [65]

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

The work input is $W = ( 245 + A ) \ J$

The heat delivered is $Q = (142 + B) \ J$

The value of A is A = 14

The value of B is B = 72

Generally the efficiency of the heat engine is mathematically represented as

$\eta = \frac{W}{Q_t}$

Here $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

$Q_t= Q + W$

=> $Q_t= 245 + A + 142 + B$

=> $Q_t= 390 + A+B$

So

$\eta = \frac{245 + A }{390 + A+ B}$

=> $\eta = 0.544$

=> $\eta = 0.544 *100$

=> $\eta = 54.4 \%$

5 0

2 years ago

What do magnets and electricity have in common

ra1l [238]

They both attract things.

3 0

2 years ago

Read 2 more answers

Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s

larisa [96]

Answer:

a) La aceleración angular es: $\alpha=2\: rad/s^{2}$

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

a)

La aceleración angular se define como:

$\alpha=\frac{\Delta \omega}{\Delta t}$

Donde:

Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
Δt es el tiempo en el que occure el cambio de velocidad angular

$\alpha=\frac{35-25}{5}$

$\alpha=2\: rad/s^{2}$

b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

$\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}$

Aqui el angulo inicial es 0, por lo tanto.

$\theta=20(5)+\frac{1}{2}(2)(5)^{2}$

$\theta=125\: rad$

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

$\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev$

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0

2 years ago