Given the data from the question, the goal kicker did not make the field goal.
To determine if the goal kicker makes the field goal, we shall determine the distance (i.e range) travelled by the ball after it was kicked by the kicker. This can be obtained as follow:
How to determine the range Initial velocity (u) = 20 m/sAngle of projection (θ) = 60 °Acceleration due to gravity (g) = 9.8 m/s²Range (R) = ?R = u²Sine(2θ) / g
R = 20² × Sine (2×60) / 9.8
R = 400 × Sine 120 / 9.8
R = 35.3 m
Since the distance travelled by the ball is 35.3 m and the goal post is situated 47 yard (43 m) away.
Thus, we can conclude that the goal kicker did not make the goal as the ball did not get to the post which is 47 yard (43 m)
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The vertical component of 9N is found to be four-thirds of the horizontal component.
a) Find the horizontal component.
b)Find the magnitude of the vector
Answer:
a) Let horizontal component be h
4/3×h= 9
4h = 9x3
h = 27/4
b) Magnitude = √ [9²+ (27/4)²
= 11.25
What are two models of light? how does each model explain part of the behavior of light?
The two most successful theories of light were the corpuscular (or particle) theory of Sir Isaac Newton and the wave theory of Christian Huygens. Newton's corpuscular theory stated that light consisted of particles that travelled in straight lines.
hope it helps
2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is
her displacement after 15.0 seconds?
Answer:
337.5m
Explanation:
Kinematics
Under constant acceleration, the kinematic equation holds:
[tex]s=\frac{1}{2}at^2+v_ot+s_o[/tex], where "s" is the position at time "t", "a" is the constant acceleration, "[tex]v_o[/tex]" is the initial velocity, and [tex]s_o[/tex] is the initial position.
Defining Displacement
Displacement is the difference in positions: [tex]s-s_o[/tex] or [tex]\Delta s[/tex]
[tex]s=\frac{1}{2}at^2+v_ot+s_o[/tex]
[tex]s-s_o=\frac{1}{2}at^2+v_ot[/tex]
[tex]\Delta s=\frac{1}{2}at^2+v_ot[/tex]
Using known information
Given that the initial velocity is zero ("skier stands at rest"), and zero times anything is zero, and zero plus anything remains unchanged, the equation simplifies further to the following:
[tex]\Delta s=\frac{1}{2}at^2+v_ot[/tex]
[tex]\Delta s=\frac{1}{2}at^2+(0)*t[/tex]
[tex]\Delta s=\frac{1}{2}at^2+0[/tex]
[tex]\Delta s=\frac{1}{2}at^2[/tex]
So, to find the displacement after 15 seconds, with a constant acceleration of 3.0 m/s², substitute the known values, and simplify:
[tex]\Delta s=\frac{1}{2}at^2[/tex]
[tex]\Delta s=\frac{1}{2}(3.0[\frac{m}{s^2}])(15.0[s])^2[/tex]
[tex]\Delta s=337.5[m][/tex]
Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q₁ and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.500 m. What is net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m-
-5.00 x 10-6 C
92
-5.00 x 10-6 C
93
0.500 m-
The answer to the question is 1.125 Newton
Formula for electrostatic force is F = ( K q1 q2 )/ r²
where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.
Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.
Electrostatic force between q1 and q3
r will be 0.500 + 0.500 = 1 m
F = ( K q1 q3 )/ r²
F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²
F₁₃ = 2.25 × 10⁻¹ N
Electrostatic force between q2 and q3
r will be 0.500 m
F = ( K q1 q3 )/ r²
F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²
F = (225 × 10⁻³) / (25 × 10⁻²)
F₂₃ = 9 × 10⁻¹ N
Total force on q3 will be F₁₃ + F₂₃
Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N
Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N
Total force on q3 = ( 11.25 × 10⁻¹ ) N
Total force on q3 = 1.125 N
Thus after solving we got the net force on q3 as 1.125 Newton
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A ball is shot from the ground straight up into the air with initial velocity of 40 ft/sec. Assuming that the air resistance can be ignored, how high does it go
The height to which the ball attained, given the data is 7.58 m
Data obtained from the questionInitial velocity (u) = 40 ft/s = 40 × 0.3048 = 12.192 m/sFinal velocity (v) = 0 m/s (at maximum height) Acceleration due to gravity (g) = 9.8 m/s²Maximum height (h) =? How to determine the maximum heightv² = u² – 2gh (since the ball is going against gravity)
0² = 12.192² – (2 × 9.8 × h)
0 = 12.192² – 19.6h
Collect like terms
0 – 12.192² = –19.6h
–12.192² = –19.6h
Divide both side by –19.6
h = –12.192² / –19.6
h = 7.58 m
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what evidence did astronomers use to prove jets travel in opposite directions?
The evidence that astronomers use to prove jets travel in opposite direction is Spectral lines from a very fast moving ionized gases.
Who are the astronomers?Astronomer is a person who study astronomy which is a scientific study that has to do with space, space bodies, comets, planets, stars and so on.
Therefore, The evidence that astronomers use to prove jets travel in opposite direction is Spectral lines from a very fast moving ionized gases.
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the 12-lead ecg shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv. what is this finding indicative of?
The 12-lead ECG test that shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, could be a diagnostic for myocardial infarction.
What is 12 - lead ECG test?
A 12-lead electrocardiogram (ECG) is a medical test that is recorded using leads, or nodes, attached to the body.
Electrocardiograms, sometimes referred to as ECGs, capture the electrical activity of the heart and transfer it to graphed paper.
When a 12-lead ECG shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, it could be a diagnostic for myocardial infarction.
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What is the change of state when magma crystallises to form igneous rock?
Answer:
liquid
Explanation:
due to high temperatures in the magma chamber all solids liquidify on reaching the surface with colder temperatures it now solidify to form igneous rock
