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3 years ago

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Physics

1 answer:

Alecsey [184]3 years ago

5 0

Time taken by the water balloon to reach the bottom will be given as

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

here we know that

$h = 18 m$

$g = 9.8 m/s^2$

now by the above formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

now in the same time interval we can say the distance moved by it will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

so it will fall at a distance 15.7 m from its initial position

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a 20-kg object travelling at 20 m/s collides head on with an 18-kg object travelling at 17 m/s.If they were locked together afte

djverab [1.8K]

Answer:

2.47 m/s

Explanation:

Momentum = Mass X Velocity

If they were locked together, it means its a perfectly inelastic collision. Therefore,

Total momentum before = Total momentum after

Total momentum before = (20 X 20) - (18 X 17)

= 94

Total momentum after = 94

Y = Object speed after collision

94 = (20+18)Y

Y = 2.47368421 m/s

3 0

2 years ago

Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as

Lesechka [4]

We are given:

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

__________________________________________________________

Momentum of Tarzan before swinging:

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

Momentum = 560 kg m/s

__________________________________________________________

Speed of Tarzan after grabbing the bananas:

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v [where v is the velocity of Tarzan]

v = 5.9 m/s

__________________________________________________________

<h3>Finding the Initial and Final KE and PE:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

Initial and Final KE:

We know that KE = 1/2*(mv²)

Initial KE:

Initial KE = 1/2*(mv²) [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

Final KE:

Final KE = 1/2*(mv²)

[where v is the velocity at the maximum point of the swing]

Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

Initial and Final PE:

We know that:

PE = mgh

[where g is the acceleration due to gravity and h is the height]

Initial PE:

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

Final PE:

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3>Finding the maximum height Tarzan will reach:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

1653.5 + 0 = 0 + 950h

1653.5 = 950h

h = 1653.5/950 [dividing both sides by 950]

h = 1.74 m

Therefore, the maximum height reached by Tarzan is 1.74 m

but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey

6 0

2 years ago

7) A crazy cat (yes, this is redundant) is running along the roof of a 60 m tall building. The cat is moving at a constant veloc

gregori [183]

Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.

The velocity vector of the cat when it reaches the ground will be:

v = (v0x, v0y + g · t)

v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)

v = (44 m/s, -34.3 m/s)

The magintude of the vector "v" is calculated as follows:

$|v| = \sqrt{(44 m/s)^{2}+(-34.3 m/s)^{2}} = 55.8 m/s$

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

6 0

2 years ago

If the mass of the block is 5 kg and the speed 7 m/s, what is the work done on the block?

Kamila [148]

Answer:

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms

2 ).

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

23 =20 3 N

While the uertical component of the force acting in upward direction=Fsin30=40× 21

=20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

N−6N=28.64N

The horizontal acceleration of the block=

Fcos30−μN = 528.64

=5.73m/s 2

8 0

3 years ago

Please answer! thank you!

lesantik [10]

Answer:

Maybe A.... I'm not sure

Explanation:

8 0

2 years ago

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