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JulijaS [17]
3 years ago
8

"the student drops one cannonball, and exactly 1.0 s later drops the other cannonball from the same height. what is the time int

erval between the first cannonball striking the ground and the second cannonball striking the ground"
Physics
1 answer:
Elden [556K]3 years ago
6 0

Answer;

1 second

Explanation;

Two objects moving at the same speed will always stay the same distance apart. If two objects are moving at different speeds, the distance between them must change.

Therefore; if the distance will be the same and the speed is also the same then the time taken will be the same.

You might be interested in
How much heat is released when 35kg of water freezes?
inn [45]
You need an additional point of data here: the enthalpy of fusion, or conversely the enthalpy of melting (they differ only by their sign). For water (or ice) that value is gotten from sources such as the internet

<span>ΔH°(fus) = 6.01 kJ/mole </span>

<span>Since you have 35 000g, how many moles do you have? </span>
<span>Moles H2O = 35000 g/(18.015 g/mole) = 1942.8 moles</span>

<span>So, take that ΔH°(fus) in kJ/mole, multiply by the number of moles, and there ya go! 
</span>
6.01 x 1942.8 = 11,676 kJ of energy is released

Hope I helped!! xx
3 0
3 years ago
a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
3 years ago
Somebody please explain how to solve this. Thanks in advance!
Jobisdone [24]

work done is product of force and displacement of point of application of force

so here we have to check the product of force and displacement both

Now we will put the least to maximum work in the following order

1. -A man exerts strenuous effort in pushing a stationary wall

2. -A flea pushes a speck of dirt 1 cm

3. -A farmer pushes a 2 kg wheelbarrow 20 m

4. -A cannon launches a 3 kg cannonball a distance of 200

5. -A 2000 kg car travels 400 m down a road

6. -Space shuttle Atlantis launches from the ground into near-Earth orbit

5 0
2 years ago
Timmy drove 2/5 of a journey at an average speed of 20 mph.
mixer [17]

Answer:

4hr

Explanation:

5 0
2 years ago
Read 2 more answers
What is the mass of a large ship that has a momentum of 1.60x10^9 kg*m/s and is moving at a velocity of 10m/s?
Elden [556K]

Answer:

160000000 kg.

Explanation:

p=mv

p=1.6x10^9

v=10m/s

rearrange and substitute:

(1.6x10^9)=m(10)

m=(1.6x10^9)/10

m= 1.6x10^8 kg.

7 0
3 years ago
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