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3 years ago

A car travelled a distance of 200 m with initial velocity of 216 km/hr. Calculate the acceleration

Physics

1 answer:

pogonyaev3 years ago

7 0

Answer:

$a = 16\ m/s^2$

Explanation:

When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

$v_f^2=v_o^2+2ax$

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

$\displaystyle a=\frac{v_f^2-v_o^2}{2x}$

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

$\displaystyle a=\frac{100^2-60^2}{2*200}$

$\displaystyle a=\frac{10000-3600}{400}$

$\displaystyle a=\frac{6400}{400}$

$\mathbf{a = 16\ m/s^2}$

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Change of motion are caused by

AlexFokin [52]

The only thing that can change the motion of an object is a net (unbalanced)force acting on it. This is given by Newton's First Law of Motion, sometimes also called the Law of Inertia.

I hope this helps!

4 0

4 years ago

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi

zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} = mass flow rate\\\rho = Density\\V = Velocity$

values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3V_1 = 40m/s$

$A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s$

PART A) Applying the flow equation

$\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2$

7 0

3 years ago

ehidna [41]

Answer:

$a = -1.6\ m/s^2$

Explanation:

Accelerated Motion

It occurs when an object changes its speed over time. If the changes in speed are uniform, then the acceleration is constant, positive if the speed increases, negative if the speed decreases.

The acceleration is calculated as follows:

$\displaystyle a=\frac{v_f-v_o}{t}$

The aeroplane starts with a speed of vo=62 m/s and reaches a speed of vf=6 m/s in t=35 s.

The acceleration is:

$\displaystyle a=\frac{6-62}{35}=-\frac{56}{35}$

$\mathbf{a = -1.6\ m/s^2}$

3 0

3 years ago

A skateboarder travels on a horizontal surface with an initial velocity of 4.0 m/s toward the south and a constant acceleration

Alenkinab [10]

Answer:

a) 0.32 m b) -2.4 m c) 1.08 m/s d) -4 m/s

Explanation:

As the x and y axes (as chosen) are perpendicular each other, the movements along these axes are independent each other.
This means that we can use the kinematic equations for displacements along both axes.
In the x direction, as the only initial velocity is in the south direction (-y axis), the skateboarder is at rest, so we can write:

$x =\frac{1}{2}*a*t^{2} (1)$

In the y-direction, as no acceleration is acting on the skateboarder, we can write the following displacement equation:

$y = v_{0y} * t (2)$

For t = 0.6s, replacing by the givens, we get the position (displacement from the origin) on the x-axis, as follows:

$x =\frac{1}{2}*a*t^{2} =\frac{1}{2} * 1.8 m/s2*(0.6s)^{2}\\ x = 0.32 m$

From (2) we can get the position on the y-axis (displacement from the origin) as follows:

$y = v_{0y} * t = -4 m/s * 0.6 s = -2.4 m$

In the x- direction, we can find the component of the velocity along this direction, as follows:

$v_{fx} = a*t$

Replacing by the values, we have:

$v_{fx} = a*t = 1.8 m/s2 * 0.6 s = 1.08 m/s$

As the skateboarder moves along the y-axis at a constant speed equal to her initial velocity, we have:

vfy = voy = -4 m/s

8 0

4 years ago

A 10 kg dog runs across the ground and jumps onto an ice-covered pond at 5 m/s. If the force of friction acting on the dog is 8

zzz [600]

Answer:

15.625m

Explanation:

Using the equation of motion

v² = u²+2as

u and v are the initial and final velocities

a is the acceleration

s is the distance

Get the acceleration

According to Newton's second law;

F = ma

8 = 10a

a = 8/10

a = 0.8m/s²

We are also given

v = 5m/s

u = 0m/s

Substitute into the formula above

v² = u²+2as

5² = 0²+2(0.8)s

25 = 1.6s

s = 25/1.6

s = 15.625m

Hence the dog will cover a distance of 15.625m before coming to rest

3 0

3 years ago