Answer:
Approximately [tex]14\; \rm m[/tex] above the height of the fire hose, assuming that air resistance is negligible and that [tex]g = -9.81\; \rm m \cdot s^{-2}[/tex].
Explanation:
Consider the motion of water particles in two directions: vertical and horizontal.
Assuming that air resistance on the water particles is negligible.
Vertically, water particles would accelerate at [tex](-9.81\; \rm m \cdot s^{-2})[/tex] (towards the ground.)Horizontally, water particles would travel at a constant velocity.Initial velocity: [tex]v_0 = 25\; \rm m \cdot s^{-1}[/tex]. Angle of elevation: [tex]\epsilon = 60^\circ[/tex]. Calculate the initial velocity of these water particles in these two components:
Initial horizontal velocity: [tex]v_0 (\text{horizontal}) = v_0 \cdot \cos\left(\epsilon\right) \approx 12.5\; \rm m \cdot s^{-1}[/tex].Initial vertical velocity: [tex]v_0 (\text{vertical}) = v_0 \cdot \sin\left(\epsilon\right) \approx 21.7\; \rm m \cdot s^{-1}[/tex].How much time would it take for a water particle reaches the building from the hose? That particle needs to travel [tex]x(\text{horizontal}) = 45\; \rm m[/tex] at a constant horizontal speed of [tex]v(\text{horizontal}) \approx 12.5\; \rm m \cdot s^{-1}[/tex]. Therefore:
[tex]\displaystyle t = \frac{x(\text{horizontal})}{v(\text{horizontal})} \approx 3.6\; \rm s[/tex].
In other words, that water particle would be approximately [tex]3.6\; \rm s[/tex] into its flight when it hits the building. What would be its height? Assuming that the air resistance on that particle is negligible. The height of that water particle at time [tex]t[/tex] may be modeled using the SUVAT equation:
[tex]\displaystyle x(\text{vertical}) = \frac{1}{2}\, a \, t^{2} + \left[v_0(\text{vertical})\right]\, t[/tex],
where:
[tex]x(\text{vertical})[/tex] gives the height of the water particle (relative to where it was launched.)[tex]a = g = -9.81\; \rm m \cdot s^{-2}[/tex] (negative because gravitational acceleration points towards the ground.)[tex]v_0(\text{vertical}) \approx 21.7\; \rm m \cdot s^{-1}[/tex] (see above.)[tex]t \approx 3.6\; \rm s[/tex] (that's the time when the water particle hits the building.)Calculate the height of the water particle when it hits the building:
[tex]\begin{aligned}x(\text{vertical}) &= \frac{1}{2}\, a \, t^{2} + \left[v_0(\text{vertical})\right]\, t \\ &\approx \frac{1}{2} \times \left(-9.81\; \rm m \cdot s^{-2}\right) \times (3.6\; \rm s)^2 + 21.7\; \rm m \cdot s^{-1} \times 3.6\; \rm s \\ &\approx 14\; \rm m\end{aligned}[/tex].
A 10.0-cm-diameter and a 20.0-cm-diameter charged ring are arranged concentrically (so they share the same axis). Assume both are centered at the origin. The 10 cm ring is charged to 30.0nC, while the 20 cm ring is charged to 60.0nC. Using the library electric field formula for Ering, compute the total electric field strength numerically at:
a, The origin
b. 40.0 cm along the axis of the 10 cm ring.
Answer:
a) Eₓ = 0 , b) c) Ex_total = 3.96 10⁻⁶ N
Explanation:
For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.
From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.
dE = k dq / r²
The component of this field in the direction to x is
d Eₓ = dE cos θ
the distance of ring to the point
r² = x² + a²
where a is the radius of each ring
the cosine is the adjacent leg between the hypotenuse
Cos θ = x /r
we substitute
d Ex = k / (x² + a²) dq x /√ (x² + a²)
we integrate
Ex = ∫ k / (x² + a²)^{3/2} x dq
Ex = k x/ (x² + a²)^{1.5} Q
the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum
Eₓ_total = Eₓ₁ + Eₓ₂
a) at the origin x = 0
in field is zero, since the numerator e makes zero
Eₓ = 0
b) for point x = 40.0cm = 0.400 m
we substitute the values in our equation
Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]
Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]
let's calculate
Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹
Ex_total = 108 [14.2668 + 22.36] 10⁻⁹
Ex_total = 3.96 10⁻⁶ N
Particles in a warmer block will have
the particles in the colder block.
kinetic energy and move
than
more / slower
less / slower
less / faster
more / faster
Answer:
more/ faster
Explanation:
I got it right on my test
Particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.
What is kinetic energy?The energy which an item has as a function of motion is determined by the initial energy. It is described as the effort required to propel a material body from rest to the given velocity.
As we know,
As the temperature rises, the particles' kinetic energy rises and they begin to move more quickly. Particles have higher kinetic energy in gases than in solids.
Thus, particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.
The question is incomplete.
The complete question is:
Particles in a warmer block will have ________ kinetic energy and move ______than the particles in the colder block.
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Please help me! I will give you a thanks!
Answer:
The answer is D
Explanation:
Because it is big and can fit the truck.
A bungee cord is 30.0 mm long and, when stretched a distance xx, it exerts a restoring force of magnitude kxkx. Your father-in-law (mass 97.0 kgkg) stands on a platform 45.0 mm above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 mm before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 420 NN.
Part A When you do this, what distance will the bungee cord that you should select have stretched?
Answer:
0.65 m
Explanation:
See attachment for calculation
a motorbike and its rider have a mass of 274kg. if they experience an acceleration of 4.50 m/s^2 what force was exerted on them
Answer:
1233 N
Explanation:
force=mass×accerelation
Pain during sports is completely normal and should not be a concern.
Please select the best answer from the choices provided.
OT
OF
Answer:
F
Explanation: Its common sence
Answer:
It’s false
Explanation:
2020 (WORST YEAR EVER) on Edge ;-)
A man throws a football straight into the air Which type of energy is the
football gaining as it rises
Answer:
It is gaining potential energy
Answer: it could also be gravitational potential energy
This is because as it rises in the air it will gain more energy to fall back down
Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical force that the objects exert on each other. Use 8.988100 N·m2/C2 for the value of Coulomb's constant. Give your answer to two decimal places.
Answer:
a) F= 0,19 [N] according to problem statement
b) F = 0,19*10⁹ [N] using the right value of K
Explanation:
The force between two electric charges is according to Coulomb´s law is:
F = K * q₁*q₂ / d² where q₁ and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant K = 8,988100*10⁹ N.m²/C². The problem establishes to use K = 8,988100 N.m²/C².
NOTE: To value of is : K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹
Then:
F = 8,988100 * 1,2* 0,36 / (4,5)² [ N*m²/C² ] * [ C*C*/m²]
F = 3,882859/ 20,25 [N]
F= 0,19 [N]
The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies
What is the mass of a 2 kg object on the Earth and on the moon?
Answer:
Same
Explanation:
Mass is the quantity of matter in a certain object.
WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.
On earth, it is 9.81 whereas on the Moon it is 1.6
Here, we are required to determine the mass of a 2kg object on the Earth and on the moon.
The mass of the 2kg object is the same on earth and on the moon.The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.
The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.In essence, if the mass of an object is 2kg, this mass remains constant on planet earth and the moon.
PS: The weight of an object which is a measure of the acceleration due to gravity which differs with location and position is the constant which may be different about the object on the earth and the moon respectively.
The acceleration due to gravity on earth and the moon is 9.8m/s² and 1.6m/s² respectively.
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A 0.10 g honeybee acquires a charge of +23 pC while flying, due to friction of the wings against the air. We will try and understand why this is advantageous. A. The electric field near the surface of the earth is typically 100 N/C, directed downward. What is the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight?
Answer:
The ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶
Explanation:
Given;
mass of the honeybee, m = 0.1 g
charge acquired by the honeybee, Q = 23pC = 23 x 10⁻¹² C
the electric field near the earth's surface, E = 100 N/C
The magnitude of the electric force on the bee is given by;
F = QE
F = (23 x 10⁻¹²)(100) = 23 x 10⁻¹⁰ N
The weight of the bee is given by;
W = mg
W = 0.1 x 10⁻³ x 9.8
W = 9.8 x 10⁻⁴ N
The the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is given by;
[tex]\frac{F}{W} = \frac{23*10^{-10}}{9.8*10^{-4}} = 2.35 *10^{-6}[/tex]
Therefore, the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶
How much force is needed to accelerate an object with a mass of 90 kg at a rate of 1.2 ms
Read each scenario and then answer the questions.
Scenario A: A shopping cart is pushed a distance of 3 m by a force of 15 N.
Scenario B: A shopping cart is pushed a distance of 4 m by a force of 12 N.
Scenario C: A shopping cart is pushed a distance of 6 m by a force of 10 N.
Which scenario requires the most work?
Which scenario requires the least work?
Answer:
C requires the most work and A requires the least work
Answer:
Which scenario requires the most work?
✔ C
Which scenario requires the least work?
✔ A
A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m. If we model her body as having a uniform cross section and assume that her center of gravity is midway between her hands and her feet, what is her initial angular acceleration?
The initial angular acceleration is 5.78 rad/s^2.
Calculation of the angular acceleration;Since
A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m.
So,
[tex]Torque \tau = r \times F\\\\ \tau = l\alpha\\\\so, r \times F = l\alpha\\\\mg\ cos\ 45 \times r = (1/3)mR^2 \times \alpha\\\\9.8\times cos\ 45 \times 0.9 = (1/3)\times (1.8)^2 \times \alpha\\\\\alpha = 5.78\ rad/s^2[/tex]
hence, The initial angular acceleration is 5.78 rad/s^2.
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A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Complete question:
A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Answer:
The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Explanation:
Given;
charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C
length of the conductor, L = 50 m
inner radius, r₁ = 1.304 mm
outer radius, r₂ = 9.249 mm
The magnitude of the electric field halfway between the two cylindrical conductors is given by;
[tex]E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}[/tex]
Where;
λ is linear charge density or charge per unit length
r is the distance halfway between the two cylindrical conductors
[tex]r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm[/tex]
The magnitude of the electric field is now given as;
[tex]E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m[/tex]
Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Use Percentages Given that the molecular mass of magnesium
hydroxide (MOH.) is 58.32 amu and the atomic mass of an atom of
oxygen is 15.999 amu, what percentage of this compound is oxygen?
Answer:
27%
Explanation:
15.999 divided by 58.32 = .27433128
Move the decimal place over 2 places.
27%
Make use of percentages. Given that the molecular mass of magnesium hydroxide (MOH.) is 58.32 amu and the atomic mass of an oxygen atom is 15.999 amu, oxygen accounts for 27% of this compound.
Molecular versions of many substancesChemical substances known as molecular compounds assume the shape of distinct molecules. Examples include common compounds like carbon dioxide (CO2) and water (H2O) (Figure 3.1.1 ).
These substances differ significantly from ionic substances like sodium chloride (NaCl). When metal atoms give up one or more of their electrons to non-metal atoms, ionic compounds are created. The cations and anions that result
Therefore,
58.32 / 15.999 (27.3%) =.27433128
Add two places to the decimal place.
27%
The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance. In light of this, the molar mass is average. Amu, short for atomic mass units, is the unit used to quantify molecular mass.
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Analysis of an accident scene indicates a car was traveling at a velocity of 70.0 mph (31.3 m/s) along the positive x-axis at the instant the brakes were applied and then traveled a distance of 115 m before hitting an obstacle at a velocity of 4.65 m/s. Determine the car's acceleration a (assumed constant) and the time t between applying the brakes and hitting the obstacle. (Assume the car is moving in the positive direction. Indicate the direction with the sign of your answer.)
Answer:
The acceleration of the car is -4.166 m/s²
The time between applying the brakes and hitting the obstacle is 6.4 s
Explanation:
Given;
initial velocity of the car when the brake was applied, u = 31.3 m/s
final velocity of the car after hitting an obstacle, v = 4.65 m/s
distance traveled before stopping, s = 115 m
The acceleration of the car is given by;
v² = u² + 2as
2as = v² - u²
a = (v² - u²) / 2s
a = (4.65² - 31.3²) / (2 x 115)
a = -4.166 m/s²
The time between applying the brakes and hitting the obstacle;
v = u + at
at = v - u
t = (v - u) / a
t = (4.65 - 31.3) / -4.166
t = 6.4 s
The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will be_______, and, consequently, the control will be better.
The acceleration along the ramp, therefore, the speed at the bottom of a slope will be:__________
a. lower
b. gcostheta
c. higher
d. gsintheta
Answer:
Lower
Lower
gsintheta (gsinθ)
Explanation:
The sum of forces resolved parallel to the inclined plane is given by;
F - mgsinθ = 0
ma - mgsinθ = 0
ma = mgsinθ
a = gsinθ
Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.
therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.
The acceleration along the ramp, is gsintheta (gsinθ)
PLS HELP
Given the following blackbody curve graph, what color will Star A be?
Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum.
Blue
Red
White
Yellow
Answer:
Yellow
Explanation:
It is for sure not red
Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum. The color will Star A be Yellow.
What is Wavelength?A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in meters (m), centimeters (cm), or millimeters (mm).
A blackbody's spectrum is continuous (it emits some light at all wavelengths), and it peaks at a particular wavelength. For hotter objects, the peak of the blackbody curve shifts to shorter wavelengths in a spectrum. If you conceive of a blackbody in terms of visible light, the hotter it is, the bluer its peak emission wavelength will be. For instance, the temperature of the sun is about 5800 Kelvin. With this temperature, a blackbody's peak is roughly 500 nanometers away, or the wavelength of yellow.
The color will Star A be Yellow.
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What is the answwr I need help
You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a constant vR equal to 34 m/s and the blue car is moving at constant vB equal to 28 m/s. At the moment you fist see them, the blue car is 29.0 m ahead of the red car.
Required:
a. How long after you first see the cars does the red car catch up to the blue car?
b. How far did the red car travel between when you fist saw it and when it caught up to the blue car?
c. Suppose the red car started to accelerate at a rate of a equal to 5/3 m/s^2 just at the moment you saw the cars.
d. How long after that would the red car catch up to the blue car?
Answer:
a) 3.66 s
b) 124.4 m
c) 3.12s
Explanation:
Given that
Speed of the Red Car, v₁ = 34 m/s
Speed of the Blue Car, v₂ = 28 m/s
Distance between the two cars, d = 22 m
The difference between the speed of the cars is: 34 - 28 = 6 m/s
From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.
1)
If we divide the distance by the difference in speed. This becomes
d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.
2
From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.
v = d * t = 34 * 3.66 = 124.4
Therefore the red car travels at 124.4 m before catching up to the blue car.
3
If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.
S = ut + ½gt², where
S = 22
u = 6
t = ?
g = 2/3
22 = 6t + 1/3t²
By using the quadratic formula, we find out the two answers listed below
t1 = 3.12 s
t2 = - 21.12 s
We all know that negative time is not possible, so the answer is t1. At 3.12 seconds
Chief Boolie,
the jungle dweller, is out hunting for dinner when a coconut falls from a
tree and lands on his toe. If the nut fell for 1.4s, how fast was it traveling
when it hit Chief Boolie's toe? *
Answer:
13.72 m/s
Explanation:
Here, we are to determine the final velocity of the coconut. Applying first equation of free fall;
v = u + gt
where: v is the final velocity of the object, u is the initial velocity, g is the force of gravity and t is the time.
Since the coconut falls in the direction of gravity, then g = 9.8 m/[tex]s^{2}[/tex].
t = 1.4 s, u = 0 m/s
So that;
v = 0 + 9.8(1.4)
= 13.72
The coconut hits Chief Boolie's toe with a velocity of 13.72 m/s
If the point B was initially at the same level as the center of the pulley C, what is the direction of the velocity of point B after the pulley at A has undergone 2 revolutions? Give this in degrees from the horizontal (positive in the counterclockwise direction, negative in the clockwise direction)
Answer:
θ = 720º
Explanation:
Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.
Let's reduce the magnitudes to the SI system
θ = 2 rev (2π rad / 1 rev) = 4π rad
we assume that the pulley has a radius r
linear and angular coordinates are related
y = θ r
let's calculate
y = 4π r
to convert the distance in angles to radians
θ = y / r
θ = 4π
let's change this angle to degrees
π rad = 180º
θ = 4π rad (180º /π rad)
θ = 720º
The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank has a density of 1.0 g/mL. How far has the boat sunk into the water?
A. 3.2 cm
B. 4.0 cm
C. 4.8 cm
D. 8.0 cm
Answer:
A that the answer I think I'm not tryna do u bad thi
Explanation:
but try it
Which method of measurement is most likely to be precise but lacks
accuracy?
A. Measuring the distance across a pond by counting how many
strokes you make as you swim across
B. Measuring time by counting the number of breaths you take as
you do an activity
O C. Measuring your weight with a scale that measures to the nearest
tenth of a kg but adds 2 kg to each measurement
D. Measuring the length of a bus using a tape measure that
measures to the nearest millimeter
Answer:
C
Explanation:
Measuring your weight with a scale that measures to the nearest
tenth of a kg but adds 2 kg to each measurement
a car with a mass of 2000kg is traveling at a velocity of 50 m/s. the engine supplies a force of 1000N forward.the force of friction is 1000 N backwards. What is the velocity after 5 seconds
Answer:
no
Explanation:
ok
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.33 F capacitor charged to 77.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 427, with one possessing a dielectric constant of 983.
A. Find the electric potential energy of the original capacitor when it is charged.
B. Calculate the electric potential energy of the upgraded capacitor when it is charged.
Answer:
A) 4.035 × 10^(9) J
B) 9.29 × 10^(9) J
Explanation:
We are given;
Capacitance of the original capacitor; C = 1.33 F
Potential difference given to the original capacitor; V = 77.9 kV = 77.9 × 10³ V
A) The formula for Potential energy (U) for the original capacitor is given as:
U = ½CV²
Plugging in the relevant values, we have;
U = ½ × 1.33 × (77.9 × 10³)²
U = 4.035 × 10^(9) J
B) We are told that the capacitor with dielectric constant of 427, was replaced with one possessing a dielectric constant of 983.
Thus;
U = ½ × 1.33 × (983/427) × (77.9 × 10³)²
U = 9.29 × 10^(9) J
3. What would happen if Earth did not rotate?
Answer:
If Earth were to stop spinning on its axis, gradually the oceans would migrate towards the poles from the equator. The atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. And this means, rocks, topsoil, trees, buildings, and so on, would be swept away into the atmosphere.
Hope that helped <3
It is the final seconds of an ice hockey game between the Flyers and the Bruins. The Bruins are down by 1 point. With 20 s left in the game, the Bruins pull the goalie and have him play as a forward in an attempt to tie the game.
The Flyers successfully defend their goal for 9 s. With only 1.25 s remaining on the game clock, a Flyer shoots the puck on the ice past the skates and sticks of the other players and toward the Bruins' goal. The puck is 37 m from the goal when it leaves the stick with an initial horizontal velocity of 30 m/s. The shot is perfectly directed toward the empty goal, but the ice slows the puck down at a constant rate of 0.50 m/s2 as it slides toward the goal. None of the Bruins can stop the puck before it reaches the goal.
a. Where is the puck when the game clock reaches zero and the horn sounds to end the game?
b. Do the Flyers win the game by 1 or 2 points?
Answer:
a
[tex]P =0.1094 \ m[/tex] past the goal post
Yes the Flyers won the match
Explanation:
Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as
[tex]D = ut + \frac{1}{2} * a * t^2[/tex]
=> [tex]D =30 * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2[/tex]
=> [tex]D =37.1094 \ m [/tex]
Generally the position of the puck when the game clock reaches zero is mathematically represented as
[tex]P = 37.1094 -37[/tex]
[tex]P =0.1094 \ m[/tex] past the goal post
Given that the Bruins where one point down and Flyers scored another goal it means that the flyer are now two point up hence they won the match
Which of the following subatomic particles determine the identify of an atom?
A. Protons
B. Neutrons
C. Electrons
D. Nucleus
LOL SOMEONE PLS HELP THIS IS PAST DUE SINCE A WEEK AGO IM STRESSING SO BAD LAMISJSKAKAO
Which change will decrease the electric force between two positively charged objects?
1.moving them closer together
2.moving them farther apart
3.adding neutrons
4.removing electrons
Answer:
2
Explanation:
Answer:
The answer is B
Explanation: