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3 years ago

A flowerpot weighing 2 kilograms, what will be its speed at the end of 5 seconds?

1 answer:

8 0

That depends on what its speed was at the beginning of 5 seconds,
and on what exactly happened to it.

If it got pushed off of the window sill and fell to the ground below,
then the answer would depend on how high the window sill is.

If the flower pot kept falling for the whole 5 seconds, then its speed
would be 49 meters per second at the end of the 5 seconds. That's
because here on Earth, gravity increases the speed of falling things
by 9.8 meters per second for every second they fall.

Naturally, this answer is only true if the window sill is more than
122.5 meters above the ground ... about 402 feet, or 40 floors.
If it's lower than that, then the flower pot hits the ground in less
than 5 seconds, and its speed at the end of 5 seconds is zero.

By the way ...

-- It makes no difference how much the flower pot weighs.
All objects fall with the same acceleration, and hit the ground
at the same time.

-- 5 kilograms is a mass, not a weight.

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In a collision, a 15 kg object moving with a velocity of 3 m/s transfers some of its momentum to a 5 kg object. What would be th

Misha Larkins [42]

The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter $p$ :

$p=mv$

The total momentum is the sum of the momentums. The initial situation is the following:

$m_A=15,\quad v_A=3,\quad m_B=5,\quad v_B=0$

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).

So, at the beginning, the total momentum is

$p=m_Av_A+m_Bv_B=15\cdot 3+5\cdot 0=45$

At the end, we have

$m_A=15,\quad v_A=1,\quad m_B=5,\quad v_B=x$

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)

After the impact, the total momentum is

$p=m_Av_A+m_Bv_B=15\cdot 1+5\cdot x=15+5x$

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for $x$ , and you'll have the final velocity of the 5-kg object.

4 0

3 years ago

Work & Power Problems

Artyom0805 [142]

Answer:

1. Unit for work is measured in Joules.

Formula for work is power * time taken

2 The unit for force is Newton

A work done in lifting a jar of water

3. The unit for measuring power is Joules.

The formula for measuring power is work done/time taken

6 0

3 years ago

Please Answer These Questions I Really Need Help Please answer these questions

wel

1. Onion cells have thick rectangular walls.
2. Both are tropospheres, the lobster and fungi have the same outer shell.
3. A bat is a mammal--it has fur, lactates and is warm-blooded. Mammals are vertebrates. Vertebrates have a dorsal nerve cord protected by bony or cartilaginous vertebrae. Arthropods are invertebrates--they do not have vertebrae. They are a specific kind of invertebrate with a jointed exoskeleton.
4, 5, 6, 7 - Sorry don't know these answers to these questions

8 0

3 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0

3 years ago

You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in

Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

Volume = $\frac{\text{mass}}{\text{density}}$

So, V = $\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}$

= $5.181 \times 10^{-8} m^{3}$

Now, we will calculate the area as follows.

Area = $\frac{\text{volume}}{\text{length}}$

= $\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}$

= $2.15 \times 10^{-11} m^{2}$

Formula to calculate the resistance is as follows.

R = $\rho \frac{l}{A}$

= $\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}$

= $2.71 \times 10^{6} ohm$

Thus, we can conclude that the resistance of given wire is $2.71 \times 10^{6} ohm$ .

4 0

3 years ago