Answer:
a
Yes it clears
b
[tex]b= 0.19 \ m[/tex]
c
No it does not clear
d
[tex]z= 0.86 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the player serves the ball is [tex]v = 23.6 \ m/s[/tex]
The height of the ball above the ground is [tex]h = 2.3 7 \ m[/tex]
The distance of the net is [tex]d = 12 \ m[/tex]
The height of the net is [tex]H = 0.9 \ m[/tex]
Generally the time taken for the ball to reach the net is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
=> [tex]t = \frac{12}{23.6}[/tex]
=> [tex]t = 0.508 \ s[/tex]
Generally the change in height of the ball after t is mathematically represented as
[tex]\Delta h = ut + \frac{1}{2} gt^2[/tex]
Here u is the initial velocity which is zero given that the ball was at rest initially
So
[tex]\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2[/tex]
=> [tex]\Delta h =1.28 \ m[/tex]
Generally the new height of the ball is mathematically evaluated as
[tex]s= h-\Delta h[/tex]
=> [tex]s = 2.37 - 1.28[/tex]
=> [tex]s = 1.09 \ m[/tex]
From the value obtained we see that [tex]s > H[/tex] hence the ball clears the net
Generally the distance between the center of the ball and the top of the net is mathematically represented as
[tex]b = s - H[/tex]
=> [tex]b = 1.09 - 0.90[/tex]
=> [tex]b= 0.19 \ m[/tex]
Given that the ball makes an angle of [tex]5^o[/tex] with the horizontal , the velocity along the x-axis is
[tex]v_x = v cos(5)[/tex]
=> [tex]v_x = 23.6 cos(5)[/tex]
=> [tex]v_x = 23.5 \ m/s[/tex]
The velocity along the y-axis is
[tex]v_y = v sin(5)[/tex]
=> [tex]v_y = 23.6 sin(5)[/tex]
=> [tex]v_y = 2.06 \ m/s[/tex]
Generally the time taken for the ball to reach the net is
[tex]t = \frac{d}{v_x}[/tex]
=> [tex]t = \frac{12}{23.5}[/tex]
=> [tex]t =0.508 \ s[/tex]
Generally the change in height of the ball after t seconds is
[tex]c = v_yt + \frac{1}{2}gt^2[/tex]
=> [tex]c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2[/tex]
=> [tex]c = 2.33[/tex]
Generally the new height of the ball after time t seconds is
[tex]e = h - c[/tex]
=> [tex]e = 2.37 - 2.33[/tex]
=> [tex]e = 0.04 \ m[/tex]
From the value obtained we see that [tex]e < H[/tex] hence the ball does not clear the net
Generally the distance between the center of the ball and the top of the net is mathematically represented as
[tex]z = H-e[/tex]
=> [tex]z = 0.90 - 0.04[/tex]
=> [tex]z= 0.86 \ m[/tex]
(a) Yes, the ball clears the net.
(b) The distance between the center of the ball and the top of the net is 0.203 m.
(c) No, the ball does not clear the net.
(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.
What is a Projectile motion?When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.
Given,
The horizontal distance traveled by the ball is 12 m.
The height of the top of the net is 0.90 m.
The height of the horizontal launch of the ball is 2.37 m.
The time for the horizontal motion of the projectile that is the ball is,
[tex]\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}[/tex]
The equation for the vertical motion of the projectile can be solved by substituting the above result.
[tex]\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}[/tex]
Now, consider the case when the ball is thrown with an angle [tex]\bold{5^{\circ}}[/tex] with the horizontal.
The horizontal component of the initial velocity is [tex]\bold{v_0 \cos 5{}^\circ.}[/tex]
The vertical component of the initial velocity is [tex]\bold{v_0 \sin 5{}^\circ.}[/tex]
For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,
[tex]\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}[/tex]
Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,
[tex]\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}[/tex]
The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.
[tex]\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}[/tex]
Thus, the distance between the center of the ball and the top of the net is -2,32 m.
When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.
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What is the normal force acting on the 7.50 kg box shown?
The answer would be 31 N. This is because Fn = 7.5*9.8*sin(25).
What is a Super Massive Black Hole?
Answer:
I dont really know much but i know that it swallow anything it comes across in space.
wassup people of this wor;d
What does the triangle mean in physics?
In physics, the triangle symbol (∆) typically represents a change or difference in a particular quantity.
A triangle is often used to denote the difference between two values of a variable. For example, if we have two values of time, t1 and t2, the change or difference in time can be represented as Δt = t2 - t1, where Δt is the triangle symbol indicating the change in time. Similarly, it can be used to represent differences in other physical quantities such as displacement (∆x), velocity (∆v), or temperature (∆T). The triangle symbol (∆) is a shorthand notation commonly used in physics to indicate changes or differences.
In terms of change in displacement, the triangle symbol (∆) represents the difference between two positions or locations. It indicates the change in the object's position from an initial point to a final point.
For example, if an object initially starts at position x1 and then moves to position x2, the change in displacement (∆x) can be calculated as ∆x = x2 - x1. Here, ∆x represents the difference or change in the object's displacement.
The magnitude of ∆x gives the overall distance traveled by the object, and its sign indicates the direction of the displacement. A positive ∆x signifies a displacement in the positive direction (e.g., to the right or upward), while a negative ∆x represents a displacement in the negative direction (e.g., to the left or downward).
Therefore, The triangle symbol (∆) is commonly used in physics equations to denote changes or differences in various quantities, including displacement, velocity, time, and more. It helps us analyze and quantify the differences between two values of a physical quantity.
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A swimmer jumps from a diving board into a pool. What would a graph of the swimmer's potential energy and total mechanical energy look like?(1 point)
Potential energy would increase, while total mechanical energy would remain constant.
Potential energy would remain constant, while total mechanical energy would increase.
Potential energy would decrease, while total mechanical energy would remain constant.
Potential energy would remain constant, while total mechanical energy would decrease.
Answer: Potential energy would decrease, while total mechanical energy would remain constant.
Explanation: I did the test and this was right.
The graph should represent the third option.
Potential energy & mechanical energy:The graph for the potential energy of the swimmer and the mechanical energy should be that there is a reduction of the potential energy and the mechanical energy should remain the same.
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Uniform Circular Motion: You need to design a wheel for testing purposes such that its rim will have an acceleration of 1.5 g when the rim is moving at 37 m/s while spinning. What should be the diameter of this wheel?
Answer:
We want to have an acceleration of:
A = 1.5*g = 1.5*9.8m/s^2 = 14.7 m/s^2
When the rim is moving at 37m/s while spinning.
Now, in circular motion we have two accelerations.
Tangential acceleration, that is the one related to the change of speed, as we have a speed of 37m/s, we can assume that is constant, then the tangential acceleration is zero.
Centripetal acceleration, is the one related to the change in direction, is perpendicular to the velocity vector and is the one that allows the circular motion.
I suppose that in this problem we want to have a centripetal acceleration of 14.7m/s^2
The equation for the centripetal acceleration is:
Ac = v^2/r
Where v = velocity and r = radius.
then we must solve:
14.7m/s^2 = (37m/s)^2/r
r = (37m/s)^2/14.7m/s^2 = 93.13 m
A piece of iron of mass 200g and tempreture 300°C is dropped into 1.00 kg of water of tempreture 20°C. Predict the final equilibrium of the water.
Answer:
The final equilibrium T_{f} = 25.7[°C]
Explanation:
In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.
That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.
The temperature of thermal equilibrium will be T_f.
The heat absorbed by water will be equal to the heat rejected by Iron.
[tex]Q_{iron} = Q_{water}[/tex]
Heat transfer can be found by means of the following equation.
[tex]Q_{iron}=m*C_{piron}*(T_{i}-T_{f})[/tex]
where:
Qiron = Iron heat transfer [kJ]
m = iron mass = 200 [g] = 0.2 [kg]
T_i = Initial temperature of the iron = 300 [°C]
T_f = final temperature [°C]
[tex]Q_{water}=m*C_{pwater}*(T_{f}-T_{iwater})[/tex]
Cp_iron = 437 [J/kg*°C]
Cp_water = 4200 [J/kg*°C]
[tex]0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C][/tex]
The pressure of liquid varies as per
its depth
Answer:
Yes.
Explanation:
The pressure varies as per the depth of the container
Analyze why force pairs do not cancel each other
Answer:
Action and reaction forces don't cancel each other out because they act on separate objects.
Explanation:
Action and reaction forces are always equal in magnitude, so it's not possible to exert more force on an object than it can exert back.
Which could most likely describe the three surfaces?
Surface 1 is ice, Surface 2 is gravel, and Surface 3
is blacktop.
Surface 1 is gravel, Surface 2 is ice, and Surface 3
is blacktop.
Surface 1 is blacktop, Surface 2 is gravel, and
Surface 3 is ice.
Surface 1 is blacktop, Surface 2 is ice, and Surface
3 is gravel.
Answer:
Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.
Explanation:
Hope this helps! :]
Answer:
C. Surface 1 is blacktop, Surface 2 is gravel, and
Surface 3 is ice.
Explanation:
A kangaroo hopped at 22 meters per second, saw a skunk and then
hopped at 52 meters per second. This is an example of which of the
following?
Acceleration
Velocity
Speed
motion graphs
Answer:
this is an example of acceleration
At which temperature could air hold the most water vapor?
O A. 24°C (75°F)
B. 2°C (36°F)
C. 35°C (95°F)
O D. 13°C (55°F)
SUBMIT
Explanation:
35 maybe hope it's right
You are working as an intern for a meteorological laboratory. You are out in the field taking measurements from a balloon that is carrying equipment designed to measure electric fields in the atmosphere. Your supervisor has asked you to determine the average volume charge density at a certain height in the air. When the balloon is at an altitude of 500 m above the ground, the electric field is measured to be 160 N/C directed downward. At 800 m above the ground, the electric field is 120 N/C downward.
(a) Determine the average volume charge density (in C/m3) in the layer of air between these two elevations. (Enter the magnitude.)
(b) Is this layer of air positively or negatively charged?
positively charged
negatively charged
Answer:
a. -6.99 × 10³⁰ C/m³ b. The layer of air is negatively charged.
Explanation:
With E₁ = electric field at 500 m above the ground = 160 N/C (it is negative since it is directed downwards).
Also, with E₂ = electric field at 800 m above the ground = 120 N/C (it is negative since it is directed downwards).
The total flux, Ψ = ∫E.dA = E₁dAcosθ + E₂dAcosθ
For the 800 m surface E is parallel to dA, that is = 0° and For the 500 m surface E is anti-parallel to dA, that is = 180°
Ψ = ∫E₁dAcos180° + ∫E₂dAcos0°
= -∫E₁dA + ∫E₂dA
= -E₁∫dA + E₂∫dA
= -E₁4πR² + E₂4πR²
= (-E₁ + E₂)4πR² where R = radius of earth = 6.4 × 10⁶ m
= (-160 N/C + 120 N/C)4π(6.4 × 10⁶ m)²
= - 40 N/C)4π(6.4 × 10⁶ m)²
= -20588.74 × 10¹² C
= -2.058874 × 10¹⁶ C
≅ -2.06 × 10¹⁶ Nm²/C.
The since charge, Q = Ψ/ε₀, the total charge through the area is thus
Q = Ψ/ε₀
= -2.06 × 10¹⁶ Nm²/C ÷ 8.854 × 10⁻¹² F/m
= -0.233 × 10²⁸ C/m²
= -2.33 × 10²⁸ C/m².
So, the charge in the volume = charge net charge of surface × width of volume. So the charge in the volume Q' = QΔh = Q(h₂ - h₁) where h₁ = 500 m and h₂ = 800 m
Q' = Q(h₂ - h₁)
= -2.33 × 10²⁸ C/m²(800 m - 500 m)
= -2.33 × 10²⁸ C/m²(300 m)
= -699 × 10²⁸ C/m³
= -6.99 × 10³⁰ C/m³
b. Since Q' = -6.99 × 10³⁰ C/m³, the layer of air is negatively charged.
How far did the football player run if it took them 30 s running at 25m/s?
Answer:
[tex]\boxed {\boxed {\sf 750 \ meters}}[/tex]
Explanation:
Distance can be found by multiplying the speed by the time.
[tex]d=s*t[/tex]
The speed is 25 meters per second. The time is 30 seconds.
[tex]s=25 \ m/s \\t= 30 \ s[/tex]
Substitute the values into the formula.
[tex]d=25 \ m/s * 30 \ s[/tex]
Multiply. Note the seconds, or "s" will cancel each other out.
[tex]d= 25 \ m * 30[/tex]
[tex]d= 750 \ m[/tex]
The football player ran 750 meters.
The ball in this activity could reach much greater speeds if not for the loss of energy in many different transformations. There is one major energy loss that is causing the ball to move more slowly, what is that energy? What force is acting upon the ball to create this loss of energy?
Please help quick!
Answer:
friction acts upon the ball when it is sliding down the slide and this action creates thermal energy
What made "Lunar Orbit Rendezvous" such an attractive method to get to the Moon that it was selected by NASA in July, 1962
Answer:
I don't know the answer
Helppppppppppp I’ll give brainliest
Answer:
[tex] \boxed{(i).... 40cm}\\ \boxed{(ii).... 18cm}\\ \boxed{(iii).... 7.5N
}\\ [/tex]
Explanation:
[tex](i).... \\ the \: weight \: of \: the \: rod \: acts \: at \: the \: center \\ which \: is \: the \: half \: of \: 80cm = 40cm \\ (ii).... \\ let \: the \: distance \: be \: \boxed{d}...then \: 22 + d = 40 \\ d = 40 - 22 = 18cm \\ (iii).... \\let \: the \: weight \: of \: the \: metal \: rod \: be \: \boxed{ w} \\ taking \: the \: moment \: about \: the \: string \to \\ 15 \times (22 - 13) = 18w \\ 18w = (9)(15) \\ 2w = 15 \\ w = 7.5N[/tex]
♨Rage♨
An applied force is a force that is applied to an object by a person or another object.
True or False
Answer:
true
Explanation:
(Need Help) (ASAP)
You are at the Grand canyon standing at the edge of a ledge 1,857 m high. you have a mass of 61 kg you decide to take a selfie to share with your science teacher when you get home witch causes you to wonder...
a. How much gravitational potential energy do you have standing on the edge of this cliff?
b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, how fast would you be feeling?
Answer: Considering the gravitational field strength being 9.8...
A) 1,120,875J
B) 191.70m/s (2 DP)
Explanation:
explained in pic
You are at the Grand canyon standing at the edge of a ledge 1,857 m high. you have a mass of 61 kg you decide to take a selfie to share with your science teacher when you get home witch causes you to wonder,
a. The gravitational potential energy you have standing on the edge of this cliff is 1,120,875J.
b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, it was felt to be fast as 191.70m/s.
What is Gravitational Potential Energy?
Gravitational potential energy is the potential energy with respect to the gravitational force. To possess gravitational potential energy, the object to be placed in a position in the gravitational field.
According to the question,
Gravitational potential energy = mgh
As given here, h = 1857 m
m = 61
Substituting the given formula,
Gravitational potential energy = 61 × 9.8 × 1875
=1,120,875 J.
Here, As Gravitation potential energy gained, the Kinetic energy lost.
So, Kinetic Energy K.E = 1,120,875 J.
As we know the formula,
K.E = 1/2 mv²
v² =K.E / (1/2 m)
= 1,120,875 × 2 / 61
v² =36,750
v = [tex]\sqrt{36750}[/tex]
= 191.702 m/s.
The speed was calculated as 191.702 m/s.
Thus,
a. The gravitational potential energy you have standing on the edge of this cliff is 1,120,875J.
b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, it was felt to be fast as 191.70m/s.
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HELP MEH QUICK PLEASE
Which of the following is a source of direct current (D/C)? *
A) a radio wave
B) a wall socket
C) a generator
D) a battery
Answer:
D) a Battery
Explanation:
The best real-life example of direct current is a battery. Batteries have positive and negative terminals on a battery, the electrons in the wires will begin to flow to produce a current.
Why is my life like this
Answer:
stop it get some help
:)
Explanation:
Answer:
My life be like oooh ahhh ooh ah
Two forces of magnitude 40N and 70N acts on an object calculate and write the resultant force acting on the object and the resultant a
direction of force when both the force act in opposite direction
A) 70N in the direction of lesser force
B) 110N,opposite direction
C) 30N,in the direction of the greater force
D) 30N in the direction of lesser force
Explanation:
Required Answer70N-40N=30N
30N in lesser forcecan somone pls help asap???!!
The correct Answer: would be A B C E
Explanation:
Can i get brainlyest ?
Desde un rascacielos de 300 m de altura se lanza un objeto con una velocidad inicial de 50 m/s. Calcula el tiempo que transcurre hasta que llega al suelo y con qué velocidad llega en cada uno de los casos: a) Si se lanza verticalmente hacia arriba. b) Si se lanza verticalmente hacia abajo. c) Si se lanza horizontalmente (En este caso calcular también la distancia al edificio cuando llega al suelo) d) Si se lanza con un ángulo de 30o (Calcular también distancia al edificio)
Answer:
a) t = 14.2 s , v = -92 m / s , b) v = - 59.16 m / s , t = 0.916 s
c) t = 7.75 s , x = 387.5 m
d) t = 10.64 s , x = 463.9 m , v = 92.2 m / s
Explanation:
This is an exercise in kinematics, suppose we take the upward direction as positive
a) is thrown up vertically.
Let's use the equation
y = y₀ + v₀ t - ½ g t²
When reaching the ground y = 0, the initial height is y₀ = 300 m and the initial velocity is v₀ = + 50m / s, to simplify we use g = 10 m /s² as the value of the acceleration of gravity, for a more exact calculation we can must use 9.80 m /s²
0 = y₀ + v₀t - ½ g t²
½ 10 t² - 50 t - 300 = 0
Let's solve the quadratic equation
t² - 10 t - 60 = 0
t = [10 ±√ (10² + 4 60)] / 2
t = [10 ± 18.4] / 2
t₁ = 14.2 s
t₂ = -4.2 s
since time must be a positive quantity, the correctors result t = 14.2 s
the speed at this point is
v = v₀ - g t
v = 50 - 10 14.2
v = -92 m / s
the sign indicates that the body is going down
b) in this case the initial velocity is vo = -50 m / s
let's calculate the velocity on the ground
v² = v₀² - 2g (y-y₀)
v² = 50 2 - 2 10 ((0- 300)
v² = 3500
v = + - 59.16 m / s
as the body is going down the correct sign is the negative
v = - 59.16 m / s
the time it takes to arrive is
v = v₀ - g t
t = (v₀ - v) / g
t = (-50 + 59.16) / 10
t = 0.916 s
c) the velocity is horizontal (vox = 50 m / s), this implies that the vertical velocity is zero voy = 0
y = y₀ + v₀ t - ½ g t²
0 = 300 + 0 - ½ 10 t²
t = √ (2 300/10)
t = 7.75 s
the horizontal displacement at this time is
x = v₀ₓ t
x = 50 7.75
x = 387.5 m
d) as it is thrown with an angle let's find each component of the velocities
v₀ₓ = v₀ cos 30
[tex]v_{oy}[/tex] = v₀ sin 30
v₀ₓ = 50 cos 30 = 43.3 m / s
v_{oy} = 50 sin 30 = 25 m / s
we look for the time of descent
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + v_{oy} t - ½ g t²
0 = 300 + 25 t - ½ 10 t²
t² - 5t - 60 = 0
we solve the second degree equation
t = [5 ±√ (5² + 4 60)] / 2
t = [5 ± 16.28] / 2
t₁ = 10.64 s
t₂ = -5.64 s
since the time must be positive the result is t = 10.64 s
the range on the x axis is
x = v₀ₓ t
x = 43.6 10.64
x = 463.9 m
the ground speed is
v_{y} = [tex]v_{oy}[/tex] - g t
v_{y} = 25 - 10 10.64
v_{y} = -81.4 m / s
speed is
v = √ (v₀ₓ² + v_{y}²)
v = √ (43.3² + 81.4²)
v = 92.2 m / s
1. A ball is thrown straight up.if the launch velocity is 15 m/s, at what velocity will the ball return to the thrower's hand?
2. A boat moves across a river going 18 m/s. At the same time there is a current flowing at a right angle to the boat at 6 m/s. What is the resulting velocity of the boat?
Answer:
1) The velocity of the ball return to the thrower's hand is -15 meters per second.
2) The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].
Explanation:
1) Let suppose that ball experiments a free fall, that is an uniform accelerated motion, in which effects from gravity and Earth's rotation can be neglected. The velocity of the ball is represented by the following equations of motion:
Position
[tex]v_{o}\cdot t -\frac{1}{2}\cdot g\cdot t^{2} = 0[/tex]
[tex]t\cdot \left(v_{o}-\frac{1}{2}\cdot g\cdot t \right) = 0[/tex] (1)
Velocity
[tex]v = v_{o}-g\cdot t[/tex] (2)
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{o}[/tex] - Initial velocity of the ball, measured in meters per second.
[tex]v[/tex] - Final velocity of the ball, measured in meters per second.
From (1), we get the time when the ball returns to the thrower's hand:
[tex]v_{o}-\frac{1}{2}\cdot g\cdot t = 0[/tex]
[tex]t = \frac{2\cdot v_{o}}{g}[/tex]
And then we apply this result in (2):
[tex]v = v_{o}-g\cdot \left(\frac{2\cdot v_{o}}{g} \right)[/tex]
[tex]v = -v_{o}[/tex] (3)
Then, the velocity of the ball return to the thrower's hand is -15 meters per second.
2) The resulting velocity of the boat ([tex]\vec v_{B}[/tex]) is represented by the vectorial sum of the velocity of the boat relative to the river ([tex]\vec v_{B/R}[/tex]) and the velocity of the river ([tex]\vec v_{R}[/tex]), both measured in meters per second, that is:
[tex]\vec v_{B} = \vec v_{R}+\vec {v}_{B/R}[/tex] (4)
If we know that [tex]\vec v_{R} = 6\,\hat{i}\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{B/R} = 18\,\hat{j}\,\left[\frac{m}{s} \right][/tex], then the resulting velocity of the boat is:
[tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex]
The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].
What is the life cycle of our sun?
Answer:
the sun is currently a mian sequence star
and will remain so for another 4-5 billion
What reaction/soccur between an acid and base.
Answer:
neutralization reaction
Explanation:
The reaction of an acid with a base is called a neutralization reaction.
es una reacción química que ocurre entre un ácido y una base produciendo sal y agua. ... Se mezcla un ácido fuerte con una base fuerte: Cuando esto sucede, la especie que quedará en disolución será la que esté en mayor cantidad respecto de la otra.
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The work done in lifting a 10 kg load of bricks to the top of a building 20 m high is (g = 9.8 ms-2
a) 98
b) 200 J
c) 980
d) 1960)
Explain the difference between mass and weight and
how they are measured.
Answer:
Mass is a measure of how much force it will take to change that path. Mass depends on how much matter – atoms and so on – there is in an object; more mass means more inertia, as there is more to get moving. ... Weight, on the other hand, is a measure of the amount of downwards force that gravity exerts on an object
There is an attachment. There is an attachment on here.
Answer:
5 i think
Explanation: