Ask question

Ask question

All categories

1 year ago

10

A 77 kg student traveling in a car with a constant velocity has a kinetic energy of 1.7 104 J. What is the speedometer reading o

f the car in km/h? answer with:_____km/h

1 answer:

AVprozaik [17]1 year ago

7 0

ANSWER

75.65 km/h

EXPLANATION

Given:

• The student's mass, m = 77 kg

,

• The kinetic energy of the student in the car, KE = 1.7 x 10⁴ J

Find:

• The speed read in the speedometer of the car, which is the speed of the student, v (in km/h)

The kinetic energy of an object with mass m, traveling at a speed v, is,

$KE=\frac{1}{2}mv^2$

Solving for v,

$v=\sqrt{\frac{2KE}{m}}$

Replace the known values and solve,

$v=\sqrt{\frac{2\cdot1.7\cdot10^4J}{77kg}}\approx21.013m/s$

Note that because the kinetic energy is given in Joules - which is equivalent to kg*m²/s², the speed we found is in m/s. Now, knowing that there are 3600 seconds in 1 hour and that 1 km is equivalent to 1000 m, we can convert this to km/s,

$v=21.013\frac{m}{s}\cdot\frac{3600s}{1h}\cdot\frac{1km}{1000m}\approx75.65km/h$

Hence, the speedometer reading of the car is 75.65 km/h, rounded to the nearest hundredth.

You might be interested in

A tiger leaps with an initial velocity of 35.0 km/hr at an angle of 13.0ᶿ with respect to the horizontal. What are the component

Lorico [155]

Answer:

Vx = 35 x cos(13deg)

Vy = 35 x sin(13deg) - gt

(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)

Explanation:

The tiger leaps up, then x and y component of its velocity are:

Vx = Vo x cos(alpha)

Vy = Vo x sin(alpha) - gt

(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)

Hope this helps!

3 0

3 years ago

A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.

mars1129 [50]

Answer:

a) v = 31.67 cm / s , b) v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

w = √ K / m

w = √ 23.2 / 0.37

w = 7,918 rad / s

a) the expression against the movement is

x = A cos (wt + Ф)

Speed is

v = dx / dt = - A w sin (wt + Ф)

The maximum speed occurs for cos = ± 1

v = A w

v = 4.0 7,918

v = 31.67 cm / s

b) as the object is released from rest

0 = -A w sin (0+ Фi)

sin Ф = 0

Ф = 0

The equation is

x = 4.0 cos (7,918 t)

v = -4.0 7,918 sin (7,918 t)

v = - 31.67 sin (7.918t)

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

7,918 t = cos⁻¹ 1.5 / 4.0

t = 1,186 / 7,918

t = 0.1498 s

We look for speed

v = -31.67 sin (7,918 0.1498)

v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

31.67 / 2 = 31.67 sin ( 7,918 t)

7.918t = sin⁻¹ 0.5

t = 0.5236 / 7.918

t = 0.06613

With this time let's look for displacement

x = 4.0 cos (7,918 0.06613)

x = 3.46 cm

6 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

A compressed spring has elastic potential energy. Please select the best answer from the choices provided T F

riadik2000 [5.3K]

Answer: True

Explanation:

As the spring is compressed, it acumulates energy, and the spring "wants to release that energy". This acumulated energy, (potential energy) is called "elastic potential energy" because of the elastical nature of the spring, that when compressed it wants to return to the original shape. So the sentence is true

6 0

2 years ago

How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position

e-lub [12.9K]

The answer is C. Final position minus initial position.

5 0

3 years ago