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svlad2 [7]
3 years ago
9

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal

l be in the air before it lands? Round your answer to the nearest tenth. s How far will the football travel before it lands? Round your answer to the nearest tenth. m
Physics
2 answers:
RUDIKE [14]3 years ago
6 0

Answer:

1.2s for the first one

15.1m for the second one on time4learning

Explanation:

Step2247 [10]3 years ago
4 0

Answer:

T = 1.2 s

T = 15.1 m = 15 m

Explanation:

This is a case of projectile motion:

TOTAL TIME OF FLIGHT:

The formula for total time of flight in projectile motion is:

T = 2 V₀ Sinθ/g

where,

T = Total Time of Flight = ?

V₀ = Launch Speed = 13.9 m/s

θ = Launch Angle = 25°

g = 9.8 m/s²

Therefore,

T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)

<u>T = 1.2 s</u>

<u></u>

RANGE OF BALL:

The formula for range in projectile motion is:

R = V₀² Sin2θ/g

where,

R = Horizontal Distance Covered by ball = ?

Therefore,

T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)

<u>T = 15.1 m = 15 m</u>

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The answer is 4.7 m for the spàthan. this is a very question! I hope this helps my friend! :-)
5 0
3 years ago
On that system is<br> In an isolated system, the net __<br> zero.
DedPeter [7]

Answer:

the net work is 0

Explanation:

in an isolated system, there is no energy entering or exiting the system, and since work is dependent on the change in energy (W = ΔE) , with no change in energy, there is no work done. the net work done on or by the system must be 0

7 0
2 years ago
Which mass is undergoing to the greatest amount of acceleration ??
lina2011 [118]

Answer:

Option (3)

Explanation:

Formula used to calculate acceleration is,

F = ma

Where F = force exerted on a mass

m = mass

a = acceleration due to force exerted on the mass

Option (1),

When F = 100 N and m = 100 kg

100 = 100a

a = 1 m per sec²

Option (2)

For F = 1 N and m = 100 kg

1 = 100a

a = \frac{1}{100}

a = 0.01 m per sec²

Option (3)

For F = 100 N and m = 1 kg

100 = 1(a)

a = 100 m per sec²

Option (4)

For F = 1 N and m = 1 kg

1 = 1(a)

a = 1 m per sec²

Therefore. acceleration in Option (3) is the maximum.

4 0
3 years ago
A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat
Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

m g =\dfrac{mv^2}{r}

9.81 =\dfrac{v^2}{0.675}

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)

where, I =\dfrac{2}{5}mr^2

\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)

0.7 u^2 + g H = 0.7 v^2 + g(2R)

0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial  speed is 3.35 m/s

8 0
2 years ago
A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)
levacccp [35]

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

5 0
2 years ago
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