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madam [21]
3 years ago
7

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 20 m/s. Assuming

air resistance is negligible, find the speed (magnitude of velocity) in meters per second of the shell at launch and 4.7 s after the launch.
Physics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer: The launch speed is 43m/s

4.7secs after launch speed is 4m/s

Explanation: To solve this we use the first equation of motion but in this case our acceleration would be -10m/s² since we are going upwards against gravity(launch).

Vf = Vi - a*t

Where Vf is the final velocity after launch, Vi is the initial velocity at launch, t is time in secs then a is acceleration

a. From the question

t = 2.3secs

Vf = 20m/s

a = -10m/s²

Substituting into the above equation we have that,

20= Vi - 10* 2.3

20 = Vi - 23

Vi = 20+23

Vi = 43 m/s

Which is the speed at launch.

b. The magnitude of speed (Vf) 4.7 sec after launch is calculated as follows using same procedure but here Vi is 43m/s as calculated

Vf = 43 - 10*4.7

Vf = 43 - 47

Vf = -4m/s

But since we are asked to find the magnitude we neglect the negative sign.

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Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,
crimeas [40]

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

4 0
3 years ago
A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a rac
shusha [124]

Answer : The correct option is (D).

Explanation :

Given that,

A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a race.

We have to find her position after she runs 45 meters.

From the attached figure,

Let A is the position of Juliet. O is the initial point such that OA = 10 m, AB = 45 m and OP = 100 m.

So, using simple mathematics, it is clear that the position of Juliet after running 45 meters will be 55 m. It is OB in the figure.

So, the correct option is (D) " 55 meters ".

3 0
3 years ago
Read 2 more answers
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
PLEASE HELP:
german

Answer:

3.39 mins.

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7 0
2 years ago
Chinese public buildings erected under a construction code of the Sung dynasty have withstood earthquakes well because the white
luda_lava [24]

Answer:

A: used has four times the tensile strength of steel and the timber frame, incorporating

Explanation:

Option A is correct because it conveys the correct message intended by the statement and has no grammatical errors.

Option B is wrong because to say "has four times the tensile strength of steel has" is just grammatically and idiomatically wrong as has is used twice in the sentence.  

Option C is wrong because the statement that has to do with the flexibility of the timber's frame is more like a separate fact and does not fall under the scope of trying to further explain a fact.

Option D is wrong because it has the same problem in Option C. The comma that is placed after "steel" breaks the sentence and hence does not provide a good understanding of why the building can withstand earthquakes.

Option E is grammatically wrong for using does in the sentence "has four times the tensile strength steel does"

8 0
3 years ago
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