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3 years ago

10

An elastic loaded balloon launcher fires balloons at an angle of [38.0o above horizontal] from the surface of the ground. if the

initial velocity is 25.0 m/s, find how far away the balloons are from the launcher when they hit the level ground again. 9)a movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. how fast must the motorcycle leave the cliff-top if it's to land on the level ground below at a distance of 90.0 m from the base of the cliff?

1 answer:

Korvikt [17]3 years ago

5 0

#1

a balloon is launched at an angle 38 degree from the horizontal with speed 25 m/s

now the two component of the velocity of balloon is given as

$v_x = vcos38$

$v_x = 25 cos38 = 19.7 m/s$

$v_y = vsin38$

$v_y = 25 sin38 = 15.4 m/s$

Now the time of flight of the motion of balloon is given as

$T = \frac{2v_y}{g}$

$T = \frac{2*15.4}{9.8}$

$T = 3.14 s$

now the horizontal distance traveled by the balloon in the given time is

$R = v_x * T$

$R = 19.7* 3.14 = 62 m$

so the balloon will land at a distance of 62 m

#9

Motor cycle moves off horizontally with speed "v"

now the height of the cliff is given as

$h = 50 m$

now time taken by the motor cycle to reach the bottom

$y = \frac{1}{2}gt^2$

$50 = \frac{1}{2}*9.8*t^2$

$t = 3.2 s$

now we know that in the given time motorcycle will cover horizontal distance 90 m

so we can say

$v* t = 90$

$v*3.2 = 90$

$v = 28.2 m/s$

so motorcycle will move off with speed 28.2 m/s

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We have: a = v/t
Here, t = 2 s [ Given ]
a = 9.8 m/s² [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

Hope this helps!

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2 years ago

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What’s an example of contact friction

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Answer:

a nightstand on a lamp table

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3 years ago

27b. You're in an airplane that flies horizontally with speed 1000 km/h (280 m/s) when an engine falls off. Ignore air resistanc

jonny [76]

Answer:

8400m

Explanation:

The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.

Therefor the horizontal distance the engine travels during its fall is

280 * 30 = 8400m

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3 years ago

A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the

Lelechka [254]

Answer:

$difference \: in \: weight = 150n - 100n = 50n$

Now,buyantant force

$difference \: in \: weight \: = volume(body) \times density \: of \: water \: \times g$

so;

$50 = {v}^{b} \times 1 \times {10}^{3} \times 9.8m {s}^{2}$

${v}^{b} = \frac{50}{1000 } \times 9.8$

$= \frac{50}{9800}$

$= 0.0051$

Now,

$mass \: in \: air \: = 150n = \frac{150}{9.8kg}$

$density = \frac{weght}{volume}$

$= \frac{150}{0.0051} \times 9.8 \\ x = 3000$

And now,

$specific \: density \: = \frac{density of \: the \: body}{density \: of \: water}$

$= \frac{3000}{1000}$

$= 3$

Hence that,specific density of a given body is 3

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3 0

2 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago