1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
galina1969 [7]
3 years ago
8

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.
Physics
1 answer:
zmey [24]3 years ago
6 0

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

     Initial velocity, u  = u sin θ

     Final velocity, v =  0 m/s    

     Acceleration = -g

     Substituting

                      v = u + at  

                      0 = u sin θ - g t

                      t=\frac{usin\theta }{g}

This is the time of flight.

Consider the horizontal motion of ball,

        Initial velocity, u =  u cos θ

        Acceleration, a =0 m/s²  

        Time, t=\frac{usin\theta }{g}  

     Substituting

                      s = ut + 0.5 at²

                      s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}

This is the range.

In this problem

              u = 30 m/s

              g = 9.81 m/s²

              θ = 45° - For maximum range

Substituting

               s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

You might be interested in
20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?
Tresset [83]
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

7 0
3 years ago
Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2
xz_007 [3.2K]

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

5 0
3 years ago
Read 2 more answers
Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.
Nikolay [14]
I think answer should be d. Please give me brainlest let me know if it’s correct
5 0
3 years ago
The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?
n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100

The measured value is 2.02 meters and the actual value is 2.00 meters.

V_{measured}=2.02\\\\V_{true}=2.00

Percent Error=\frac{2.02-2.00}{2.00} *100

First, evaluate the fraction. Subtract 2.00 from 2.02

Percent Error=\frac{0.02}{2.00}*100

Next, divide 0.02 by 2.00

PercentError=0.01 *100

Finally, multiply 0.01 and 100.

Percent  Error=1\\Percent  Error= 1 \%

The percent error is 1%.

6 0
3 years ago
Other questions:
  • Why not two magnetic field lines can intersect? why not it is possible?
    13·1 answer
  • All electric circuits have devices that are run by _____ energy.
    14·1 answer
  • At one instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The car has a greater veloc
    7·1 answer
  • How does genetic drift affect small populations differently than large populations?
    9·1 answer
  • A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch
    10·1 answer
  • A cup falls off of a table from a height of 0.75 m. What is the impact speed of the cup?
    5·2 answers
  • One advantage of the __________ model is quick recognition. A. prototype B. exemplar C. language D. concept Please select the be
    9·1 answer
  • ⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
    11·1 answer
  • the rock of 10 kg is falling near the Earth's surface.assume that g =10N/kg and the is no air resistance. what is the accelerati
    5·1 answer
  • . a boat can travel in still water. (a) if the boat points directly across a stream whose current is what is the velocity (magni
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!