Show that for a projectile d2 (v2) / dt2 = 2g2

Car A rounds a curve of 150‐m radius at a constant speed of 54 km/h. At the instant represented, car B is moving at 81 km/h but

Snezhnost [94]

Answer:

Velocity of Afrom B=21m/s

Acceleration of A from B=1.68m/s°2

Explanation:

Given

Radius r=150m

Velocity of a Va= 54km/hr

Va=54*1000/3600=15m/s

Velocity of b Vb=82km/hr

VB=81*1000/3600=22.5mls

The velocity of Car A as observed from B is VBA

VB= VA+VBA

Resolving the vector into X and Y components

For X component= 15cos60=7.5m/s

Y component=22 5sin60=19.48m/s

VBA= √(X^2+Y^2)

VBA= ✓(7.5^2+19.48^2)=21m/s

For acceleration of A observed from B

A=VA^2/r= 15^2/150=1.5m/s

Resolving into Xcomponent=1.5cos60=0.75m/s

Y component=3cos60=1.5

Acceleration BA=√(0.75^2+1.5^2)

1.68m/s

4 0

2 years ago

Read 2 more answers

An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o

Karo-lina-s [1.5K]

Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

4 0

2 years ago

Sebuah benda dijatuhkan bebas dari ketinggian 200 m jika grafitasi setempat 10 m/s maka hitunglah kecepatan dan ketinggian benda

Pie

Please post in English so i or someone else can help you.

7 0

2 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

2 years ago

A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat

Zinaida [17]

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

$v=u+at$ ----------1

$s=ut+\dfrac{1}{2}at^2$ ------2

From equation 1 and 2

$s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )$

So we can say that

$s= \left(\dfrac{v+u}{2}\right)t$

Given that the velocity of car at final condition will be zero (v=0)

$s= \left(\dfrac{0+u}{2}\right)t$

$s= \left(\dfrac{u}{2}\right)t$

From the above equation we can find the initial velocity of car without finding the acceleration

$20= \dfrac{u}{2}\times 4$

u = 10 m/s

7 0

2 years ago