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oksian1 [2.3K]
3 years ago
5

You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o

f 65.0° above the horizontal. If it hits the top of the wall at a speed of 16.3 m/s, how high is the wall?
Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer:

The value is  h  =  13.2 \  m

Explanation:

From the question we are told that

    The speed of the rope with hook is u =  24 .1 \  m/s

     The angle is  \theta = 65.0^o

      The speed at which it hits top of the wall is  v  =  16.3 m/s

Generally from kinematic equation we have that

      v_y^2  =  u_y ^2 + *  2 (-g)* h

Here h is the height of the wall so

      [16.3 sin (65)]^2 =  [24.1 sin (65)] ^2+   2 (-9.8)* h

=>    h  =  13.2 \  m

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A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w
inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

6 0
3 years ago
A bike with 15cm diameter wheels accelerates uniformly from rest to a speed of 7.1m/s over a distance of 35.4m. Determine the an
Finger [1]

Answer:

9.47 rad/s^2

Explanation:

Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,

s = 35.4 m

let a be the linear acceleration.

Use III equation of motion.

v^2 = u^2 + 2 a s

7.1 x 7.1 = 0 + 2 x a x 35.4

a = 0.71 m/s^2

Now the relation between linear acceleration and angular acceleration is

a = r x α

where,  α is angular acceleration

α = 0.71 / 0.075 = 9.47 rad/s^2

8 0
3 years ago
a 3520 kg truck moving north at 18.5 m/s makes an INELASTIC collision with an 1480 kg car moving east after colliding they have
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Answer:

Explanation:

An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is

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