Answer:
Equilibrium temperature will be [tex]T=52.2684^{\circ}C[/tex]
Explanation:
We have given weight of the lead m = 2.61 gram
Let the final temperature is T
Specific heat of the lead c = 0.128
Initial temperature of the lead = 11°C
So heat gain by the lead = 2.61×0.128×(T-11°C)
Mass of the water m = 7.67 gram
Specific heat = 4.184
Temperature of the water = 52.6°C
So heat lost by water = 7.67×4.184×(T-52.6)
We know that heat lost = heat gained
So [tex]2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)[/tex]
[tex]0.334T-3.67=1688-32.031T[/tex]
[tex]T=52.2684^{\circ}C[/tex]
By equating the heat gained by the lead to the heat lost by the water, the final temperature is calculated to be approximately 41.5°C.
To determine the final temperature of the lead weight and water, we can use the principle of calorimetry, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.
Given:
Mass of lead[tex](m_l_e_a_d) = 2.61 g[/tex]
Specific heat capacity of lead [tex](c_l_e_a_d)[/tex] = 0.128 J/g°C
Initial temperature of lead[tex](T_l_e_a_d_i_n_i_t_i_a_l)[/tex]) = 11.1°C
Mass of water [tex](m_w_a_t_e_r)[/tex]= 7.67 g
Specific heat capacity of water[tex](c_w_a_t_e_r)[/tex] = 4.184 J/g°C
Initial temperature of water [tex](T_w_a_t_e_r_i_n_i_t_i_a_l)[/tex] = 52.6°C
Let T_final be the final equilibrium temperature.
Heat gained by lead [tex](Q_lead) = m_lead * c_lead * (T_final - T_lead_initia[/tex]l)
Heat lost by water [tex](Q_water) = m_water * c_water * (T_water_initial - T_final)[/tex]
Since the system is thermally insulated, Q_lead = -Q_water
[tex]m_lead * c_lead * (T_final - T_lead_initial) = - m_water * c_water * (T_water_initial - T_final)[/tex]
Solving for T_final:
[tex]T_final = (m_lead * c_lead * T_lead_initial + m_water * c_water * T_water_initial) / (m_lead * c_lead + m_water * c_water)[/tex]
Plugging in the values:
T_final = (2.61 g * 0.128 J/g°C * 11.1°C + 7.67 g * 4.184 J/g°C * 52.6°C) / (2.61 g * 0.128 J/g°C + 7.67 g * 4.184 J/g°C)
T_final ≈ 41.5°C
Therefore, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 41.5°C.
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1.67 points When comparing equal volumes of gases at the same pressure and temperature, different gases have different densities. Which property or properties of gas particles contribute to different gases having different macroscopic densities
Answer:
Molar mass due to different atomic masses
Explanation:
Two gases that have the same volume, pressure and gases have the same number of moles, as we can deduce from the ideal gas equation. So the gases will have the same number of moles of gases but they can have densities due to the fact that they have different molar masses due to the fact that they have different atomic masses. So one gas will be heavier than the other, for the same volume.
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 8.0 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point. At what horizontal distance left of the hoop should you launch?
Answer:
x = 7.62 m
Explanation:
First we need to calculate the weight of the rocket:
W = mg
we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:
W = 0.5 * 9.8 = 4.9 N
We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:
F - W = m*a
a = F - W/m
Solving for a:
a = (8 - 4.9) / 0.5
a = 6.2 m/s²
As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:
y = vo*t + 1/2 at²
y = 1/2at²
Solving for t:
t = √2y/a
t = √2 * 20 / 6.2
t = √6.45 = 2.54 s
Now that we have the time, we can calculate the horizontal distance:
x = V*t
Solving for x:
x = 3 * 2.54 = 7.62 m
People fighting forest fires carry emergency tents that haveshiny aluminum outer surfaces. If there is trouble, a fire fightercan lie under the tent to block the heat from burning treesoverhead. The tent helps because
A) radiation carries heat downward toward the fire fighter and thealuminum tent reflects most of that radiation.
B) convection carries heat downward toward the fire fighter and thealuminum tent blocks most of the heat carried by convection.
C) both conduction and radiation carry heat downward toward thefire fighter and the aluminum tent blocks most of that heat.
D) conduction carries heat downward toward the fire fighter and thealuminum tent conducts that heat harmlessly into the ground.
Answer:
option A
Explanation:
The correct answer is option A
People fighting forest fire carry emergency tent which has a shiny outer surface because the radiation of the heat which is downward toward firefighter the aluminium tents reflects the heat from the fire.
Aluminium Reflects 95 % of the infrared heat that hit the shiny surface of aluminium i.e. it is used by the firefighter to prevent from the heat radiation.
Final answer:
Emergency tents with shiny aluminum outer surfaces protect firefighters from heat primarily by reflecting infrared radiation. Convection plays a lesser role as hot air rises, and conduction is minimal. Radiative heat, the main form of heat transfer from fire, is what the tents are designed to protect against.
Explanation:
The reason emergency tents with shiny aluminum outer surfaces help firefighters when they lie under them to block heat from burning trees overhead is due to the tent's ability to reflect infrared radiation. Heat from fires is primarily transferred through radiation, particularly infrared radiation, which the shiny aluminum surface is effective at reflecting away. This reflection helps to protect the firefighter by reducing the amount of heat that reaches them. Convection plays a smaller role in transferring heat downward in this scenario, since hot air tends to rise, and conduction is negligibly slow here, especially from the fire to the tent. Therefore, the tents are designed to reflect the intense radiative heat that a fire emits, which is the main mechanism at play in this scenario for heat transfer.
A 2 kg object moves in a circle of radius 4 m at a constant speed of 3 m/s. What is the angular momentum of the object with respect to an axis perpendicular to the circle and through its center? 7. (A) 9 Nm/kg.
(B) 12 m2/s.
(C) 13.5 kg m2/s2.
(D) 18 N m /kg.
(E) 24 kg m2/s
Answer:
The angular momentum in this case is [tex] \mathbf{24\,\frac{kg\,m^{2}}{s}} [/tex]
Explanation:
The angular momentum of a point mass moving around an axis of rotation is the cross product between the distance of the object to the axis (r) of rotation and the linear momentum (p) of the particle:
[tex] \overrightarrow{L}=\overrightarrow{r}\times\overrightarrow{p} [/tex] (1)
But linear momentum is defined as mv, so (1) is:
[tex]\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v} [/tex](2)
And its magnitude is:
[tex]L=rmv*\sin\theta=(4)(2)(3)\sin(90)=\mathbf{24\,\frac{kg\,m^{2}}{s}} [/tex] (3)
It is important to note that [tex] \theta [/tex] is the angle between the radius vector and the velocity vector, because the axis of rotation is perpendicular to the circle and through its center this angle is equal to 90° and [tex] \sin(90) = 1 [/tex]
A turntable is spinning on a frictionless axle when a ball of putty is dropped onto it from above. The putty sticks to the turntable when it lands. Is angular momentum conserved for the turntable and putty (the system) during this process?
Answer:
Yes
Explanation:
Gravity and forces form the axle (external forces) do not cause forces that tend to cause rotation around the turntable's axle.
how many newtons of force will 2 pieces of paper have if one has a charge of .2 Coilombs and the other has a charge of .3 Coulombs and they are .005 meters apart?
What will happen if they are the same charge (+ or - ) Opposite charges ?
1) The electrostatic force between the two pieces of paper is [tex]2.16\cdot 10^{13} N[/tex]
2)
If the charges of the two pieces of paper are both + or both -, they repel each otherIf the charges of one piece of paper is positive and the other one is negative, they attract each otherExplanation:
1)
The electrostatic force between two charged objects is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem, we have the following situation:
[tex]q_1 = 0.2 C[/tex] is the charge on the first piece of paper
[tex]q_2 = 0.3 C[/tex] is the charge on the second piece of paper
[tex]r=0.005 m[/tex] is their separation
Substituting into the equation, we find the magnitude of the electrostatic force between them:
[tex]F=k\frac{q_1 q_2}{r^2}=(8.99\cdot 10^9) \frac{(0.2)(0.3)}{(0.005)^2}=2.16\cdot 10^{13} N[/tex]
2)
The electrostatic force can be either attractive of repulsive, depending on the relative sign of the charges of the objects involved.
In particular, we have:
If the two charges have same sign (both positive or both negative), the force between them is repulsiveIf the two charges have opposite sign (one positive and one negative), the force between them is attractiveTherefore, in this case:
If the charges of the two pieces of paper are both + or both -, they repel each otherIf the charges of one piece of paper is positive and the other one is negative, they attract each otherLearn more about electric force:
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A spring with k = 53 N/m hangs vertically next to a ruler. The end of the spring is next to the 18 cm mark on the ruler. If a 2.4 kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with on the ruler marks when the mass is at its lowest position?
Answer:
1.07 m
Explanation:
x = Compression of the spring
k = Spring constant = 53 N/m
Initial length = 18 cm
P = Kinetic energy
K = Kinetic energy
At the lowest point of the mass the energy conservation is as follows
[tex]P_{ig}+P_{is}+K_i=P_{fg}+P_{fs}+K_f\\\Rightarrow mgx+0+0=mgx+\frac{1}{2}kx^2\\\Rightarrow x=\frac{2mg}{k}\\\Rightarrow x=\frac{2\times 2.4\times 9.81}{53}\\\Rightarrow x=0.89\ m[/tex]
At its lowest position the mark on the ruler will be
[tex]x_f=0.18+0.89\\\Rightarrow x_f=1.07\ m[/tex]
The spring line will end up at 1.07 m
In this scenario, when a 2.4 kg mass is attached to a spring, the spring gets displaced due to the force exerted by the mass due to gravity. The displacement can be calculated using Hooke's Law (F = kx), resulting in a mark of 62.4 cm on the ruler when the mass is at its lowest position.
Explanation:
The phenomenon described in this question is related to Hooke's Law of Physics, which states that the force required to extend or compress a spring by a distance is proportional to that distance. The force exerted by the spring is measured in Newtons (N) and is given by the equation F = kx, where k is the spring constant and x is the distance the spring is stretched or compressed.
In this case, when the 2.4 kg mass is attached to the spring, it will exert a force due to gravity which is equal to the mass times the acceleration due to gravity (g = 9.8 m/s²), therefore F = m*g = 2.4kg * 9.8m/s² = 23.52 N. The spring reacts to this force and gets displaced, which can be calculated using x = F/k = 23.52N / 53N/m = 0.444 m or 44.4 cm. Therefore, with the mass attached, the bottom of the spring would fall to 18cm + 44.4 cm = 62.4 cm on the ruler.
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A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 50.0 s. The distance between two successive crests is 32 m. Determine, if possible, the wave's
(a) period.
(b) frequency.
(c) wavelength.
(d) speed.
(e) amplitude.
If it is not possible to determine any of these quantities, then so state.
Answer
given,
number of crest (N)= 5
time(t) = 50 s
distance between to successive crest = 32 m
a) Period
[tex]T = \dfrac{t}{N}[/tex]
[tex]T = \dfrac{50}{5}[/tex]
[tex]T = 10\ s[/tex]
b) frequency
[tex]f = \dfrac{1}{T}[/tex]
[tex]f = \dfrac{1}{10}[/tex]
[tex]f =0.1\ Hz[/tex]
c) wavelength
distance between to consecutive crest is wavelength
wavelength = 32 m
d) speed
v = f λ
v = 0.1 x 32
v = 3.2 m/s
e) Amplitude
We cannot determine amplitude because data is not given.
The period of the wave is 10.0 seconds, frequency is 0.1 Hz, wavelength is 32 meters, and speed is 3.2 m/s. The amplitude cannot be determined with the given data.
We can determine multiple properties of the wave based on the given information. Here are the steps and corresponding solutions:
Period (T): Period refers to the time it takes for one complete wave cycle to pass a point. With 5 wave crests passing in 50.0 s, it took 50.0 s / 5 = 10.0 s for each crest to pass. Therefore, the wave period is 10.0 s.Frequency (f): Frequency is the number of wave crests passing a certain point per unit time. It is the inverse of the period. So, f = 1 / T = 1 / 10.0 s = 0.1 Hz. Therefore, the wave frequency is 0.1 Hz.Wavelength (">">λ"): The wavelength is the distance between successive crests, which is directly given as 32 m. Therefore, the wavelength is 32 m.Speed (v): The speed of a wave is calculated by the formula v = f imes λ. Substituting the known values, v = 0.1 Hz imes 32 m = 3.2 m/s. Therefore, the wave speed is 3.2 m/s.Amplitude (A): Amplitude is the maximum height of the wave crest above its equilibrium position. From the provided data, it is not possible to determine the amplitude directly.
What was: The Big Bang (Science)
ANSWER FOR 10 POINTS!!
Answer:The universe sprang into existence as singularity around 13.7 billion years ago.
Explanation:
On the moon the surface temperature ranges from 379 K during the day to 1.04 x 102 K at night. Convert these temperatures to the Celsius and Fahrenheit scales.
(a) 379 K in degrees Celsius;
(b) 379 K in degrees Fahrenheit;
(c) 1.04 x 102 K in degrees Celsius;
(d) 1.04 x 102 K in degrees Fahrenheit.
Answer:
Explanation:
Relation between Celsius and Kelvin is
K = C + 273
Relation between Kelvin and Fahrenheit is
[tex]\frac{K-273}{100}=\frac{F -32}{180}[/tex]
(a) 379 K
Relation between Celsius and Kelvin is
K = C + 273
So, C = K - 273 = 379 - 273 = 106
Thus, 379 K = 106°C.
(b) 379 K
Relation between Kelvin and Fahrenheit is
[tex]\frac{K-273}{100}=\frac{F -32}{180}[/tex]
[tex]\frac{379-273}{100}=\frac{F -32}{180}[/tex]
F = 222.8°F
Thus, 379 K = 222.8°F
(c) 1.04 x 10^2 K = 104 K
Relation between Celsius and Kelvin is
K = C + 273
So, C = K - 273 = 104 - 273 = -169
Thus, 104 K = - 169°C.
(d) 1.04 x 10^2 K = 104 K
Relation between Kelvin and Fahrenheit is
[tex]\frac{K-273}{100}=\frac{F -32}{180}[/tex]
[tex]\frac{104-273}{100}=\frac{F -32}{180}[/tex]
F =- 272.2°F
Thus, 104 K = - 272.2°F
Which is the best aerobic exercise plan?
Answer:
Running or Jogging
Running and jogging are both great options for aerobic conditioning. Whether you run at the gym or outside, you are in control of setting the intensity of your workout. When aiming to build muscle mass, you can add more resistance or jog at an incline, along with increasing your speed.
Explanation:
Answer:
Running
Jogging
Yoga
Explanation:
Running Jogging and yoga best describes aerobic exercise plan. Aerobic exercises are those where cardiovascular conditioning happen it can be swimming, cycling ,running, jogging etc. aerobic exercise is good for health as it reduces risk of:- obesity, high blood pressure type 2, diabetes, metabolic syndrome , strokes and some type of cancer too can be avoided by doing aerobic exercise.
A wheel rotates with a constant angular speed ω. Which one of the following is true concerning the angularacceleration α of the wheel, the tangential accelerationaT of a point on the rim of the wheel, and thecentripetal acceleration ac of a point on the rim?
a)α= 0rad/s2, aT= 0m/s2, ac= 0 m/s2
b)α=0 aT0 ac= 0
c) α0 aT=0 ac=0
d)α=0 aT=0 ac0
e)α 0 aT0 ac0
Answer:
None of the given options
Explanation:
If the angular speed is constant, there is no angular acceleration:
α=0
If there is no angular acceleration, tangential acceleration will also be 0.
Centripetal acceleration, on the other hand, will be:
[tex]ac =\omega^2*R[/tex] (Different to 0)
An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting point to cool the coffee. What is the temperature of the coffee once the ice has melted and the system is in thermal equilibrium? Treat the coffee as though it were pure water. (Answer in °C)
Answer:
the final temperature is T f = 64.977 ° C≈ 65°C
Explanation:
Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:
Q coffee + Q ice = Q surroundings =0 (insulated)
We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).
The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)
The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)
therefore
m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0
assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water
d co = dw = 1 gr/cm³
therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr
m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0
m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L
T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )
replacing values
T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C
T f = 64.977 ° C
A small car traveling with 5,000 kgm/s hits a light truck that was at rest at a red light. After the collision the small car bounces backwards with a momentum of 3,000 kgm/s. What would the momentum of the truck be after the collision?
Answer:
8000
Explanation:
cause in the problem we dont have any ext Forces we know that the momentum of system is constant.
Fext=0
P1 + P2 = P1' + P2'
P2 = 0
5000 = -3000 + P2'
P2' = 5000+3000 = 8000
While painting the top of an antenna 275 m in height, a worker accidentally lets a 1.00 L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break.
Answer:
The force of impact of the water bottle is F = 13,475 N
Explanation:
Given data,
The height of the antenna, h = 275 m
The mass of the 1 L water bottle, m = 1 kg
Let the bottle moves distance immediately after the impact is, d = 0.2 m
The force exerted by the bottle on the bushes at the ground is given by the formula,
F = mgh / d
Substituting the values
F = 1 x 9.8 x 275 / 0.2
= 13,475 N
The value of the force of impact can be reduced by increasing the value of d, it is like the lowering the hand along with the motion of the ball to catch it thereby reduce the force of impact.
The force of impact of the water bottle is F = 13,475 N
Humans can see several thousand shades of color but have cone photoreceptors that are sensitive to only three (perhaps four) wavelengths of light. What is the best explanation for why we see so many colors?
Select one:
a.Color perception is dependent on the millions of rods as well as cone photoreceptors.
b.Color perception is achieved by activation of various combinations between the three cone types.
c.Shades of color are purely psychological and learned by association with age, infants only seeing in black and white. Incorrect
d. Colors are added and enhanced in the primary visual cortex of the brain.
Answer:
B. Color perception is achieved by activation of various combinations between the tree cone types.
Explanation:
Human eye have only three photo receptor ( three wavelength) but they can see several thousands of shades of color because color perception is gained by combination of these three wave length cone types ( RED, BLUE AND GREEN) into various combinations. Our retina have two types of cell:- cone cells and cylindrical cells. Cone cells are responsible for the identification of color.
A vessel at rest explodes, breaking into three pieces. two pieces, having equal mass, fly off perpendicular to one another with the same speed of 60 m/s. One goes along the negative x-axis and the other along the negative y-axis. The third piece has three times the mass of each other piece. What are the direction and magnitude of its velocity immediately after the explosion? What is the angle with respect to the x-axis? What is the magnitude of its velocity?
Answer:
v₃ = 28.2842 m/s
∅ = 45°
Explanation:
Given info
vi = 0 m/s
m₁ = m₂ = m
m₃ = 3m
mi = m₁ + m₂ + m₃ = m + m + 3m = 5m
v₁x= - 60 m/s
v₁y= 0 m/s
v₂x= 0 m/s
v₂y= - 60 m/s
We can apply the Principle of Conservation of Momentum as follows
pix = pfx ⇒ mi*vix = m₁*v₁x + m₂v₂x + m₃*v₃x
⇒ 5m*(0) = m*(-60) + m*(0) + 3m*v₃x
⇒ 0 = -60*m + 3m*v₃x ⇒ v₃x = 20 m/s (→) (I)
piy = pfy ⇒ mi*viy = m₁*v₁y + m₂v₂y + m₃*v₃y
⇒ 5m*(0) = m*(0) + m*(-60) + 3m*v₃y
⇒ 0 = -60*m + 3m*v₃y ⇒ v₃y = 20 m/s (↑) (II)
then
v₃ = √(v₃x² + v₃y²)
⇒ v₃ = √((20 m/s)² + (20 m/s)²) = 20√2 m/s = 28.2842 m/s
is the magnitude of its velocity immediately after the explosion
∅ = tan⁻¹(v₃y / v₃x)
⇒ ∅ = tan⁻¹(20 m/s / 20 m/s) = tan⁻¹(1) = 45°
is the direction of its velocity immediately after the explosion (the angle with respect to the x-axis)
The third piece of the exploding vessel moves along the positive x and y axes due to the conservation of momentum. The magnitude of its velocity can be derived by applying the Pythagorean theorem to its momentum components and its angle with respect to the x-axis is 45 degrees.
Explanation:This question involves concepts from physics specifically dealing with the conservation of momentum in two-dimensional motion. As the vessel explodes, the total initial momentum of the vessel (at rest) should be equal to the total final momentum, given that no external force is acting on the system.
Since the two pieces with equal mass fly off perpendicular to each other at the same speed, the x and y-components of their momentum will cancel each other out as they are equal in magnitude but opposite in direction. Consequently, the third piece with thrice the mass must account for the total momentum. Therefore, its direction of motion will be along the positive x and y axes.
To figure out the magnitude of velocity, we use the Pythagorean theorem on the x and y momentum components (which are equal in this scenario). We then derive the velocity of the last piece by the ratio of its momentum to its mass. The angle with respect to the x-axis would be 45 degrees as it is moving equally along the positive x and y axes.
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A tennis ball bouncing on a hard surface compresses and then rebounds. The details of the rebound are specified in tennis regulations. Tennis balls, to be acceptable for tournament play, must have a mass of 57.5 g. When dropped from a height of 2.5 m onto a concrete surface, a ball must rebound to a height of 1.4 m. During impact, the ball compresses by approximately 6 mm.
How fast is the ball moving when it hits the concrete surface? (Ignore air resistance.)
The velocity of the ball is 7 m/s
Explanation:
The motion of the ball is a free fall motion, so it means that the ball falls down under the effect of the force of gravity only. Therefore, it has a constant acceleration (acceleration of gravity, g), and we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the ball in this problem, we have:
u = 0 (initial velocity, the ball is dropped from rest)
[tex]a=g=9.8 m/s^2[/tex] (acceleration of gravity)
s = 2.5 m (vertical displacement)
Solving for v, we find the velocity at which the ball hits the concrete surface:
[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(2.5)}=7 m/s[/tex]
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A mountain biker encounters a jump on a race course that sends him into the air at 35,2° to the horizontal. If he lands at a horizontal distance of 30,1 m and 14,7 m below his launch point, what is his initial speed?
The initial speed of the mountain biker can be determined using principles of projectile motion, by first calculating the time of flight from the vertical displacement and then substitifying this into the equation for horizontal displacement.
Explanation:This problem can be solved using principles of projectile motion, where motion is analysed separately along the vertical and horizontal axes.
The time of flight can be determined by considering the vertical displacement and using the equation y = V0y*t - 0.5*g*t^2, where y is the vertical displacement, V0y is the initial vertical velocity, t is the total time, and g is the acceleration due to gravity. Solving for t, we can substitute this into the equation for horizontal displacement x = V0x*t, where x is the horizontal displacement and V0x is the initial horizontal velocity. This allows us to solve for the initial speed.
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The mountain biker's initial speed is found using the principles of projectile motion by resolving into horizontal and vertical components and solving a quadratic equation. The initial speed needed to achieve the jump is approximately 13.65 m/s.
Determining the Initial Speed of the Mountain Biker
To find the initial speed of the mountain biker, we'll use the principles of projectile motion. We'll perform a step-by-step analysis considering horizontal and vertical components separately.
Given data:
Angle of projection: 35.2°Horizontal distance: 30.1 mVertical distance: -14.7 m (below launch point)Step-by-Step Solution
Step 1: Separate the initial speed into horizontal (Vx) and vertical (Vy) components:
[tex]V_x = V_0 \cos(35.2^o) \\V_y = V_0 \sin(35.2^o)[/tex]
Step 2: Use the horizontal motion formula to express time (t):
[tex]x=V_xt[/tex]
[tex]$\begin{equation}t = \frac{x}{Vx} = \frac{30.1}{V_0 \cos(35.2^o)}\end{equation}$[/tex]
Step 3: Use the vertical motion formula considering the displacement:
[tex]$\begin{equation}y = V_y t + \frac{1}{2} a t^2\end{equation}$[/tex]
Substituting known values and simplifying, we have:
[tex]$\begin{equation} -14.7 = (V_0 \sin(35.2^o)) \times \frac{30.1}{V_0 \cos(35.2^o)} + \frac{1}{2}(-9.8) \left( \frac{30.1}{V_0 \cos(35.2^o)}\right)^2\end{equation}$[/tex]
[tex]or, -14.7=V_0\times0.57\times \frac{30.1}{V_0\times 0.817} -4.9\times (\frac{30.1}{V_0\times 0.817})^2 \\or,-14.7=20.99-4.9\times\frac{906.01}{V_0^2\times0.66} \\or, 6650.9/V_0^2=20.99+14.7\\or,V_0^2=6650.9/35.69\\or, V_0=13.65m/sec[/tex]
The initial speed is 13.65 m/sec.
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support. To what distance x can a person who weighs 522 N walk on the overhanging part of the plank before it just begins to tip?
Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
Technician A says that for best results, loosen the fastener immediately after spraying it with penetrating oil. Technician B says that rusted fasteners may be loosened by tapping with a hammer. Which technician is correct?
a. technician A only
b. technician B only
c. both technicians A and B
d. neither technician A nor B
Answer:
C. Both Technicians A and B
Explanation:
To loosen a fastener immediately especially if it is a rusted one we can follow these steps.
1. Tap the sides of the fastener so that the rust particle fall off.
2. These rust particles can be brushed off.
3. Spray the fastener with the penetrating oil.
4. Again tap the sides with a hammer to let the oil penetrate the inter-locked parts (threads in case of a nut and bolt).
5. Loosen the fastener after oil has penetrated.
Note: It should be kept in consideration that spraying alone won't do the job, tapping is essential to let the oil penetrate deep in to free the rusted parts.
Duplain st. is 300m long and runs from west to east between Baron and Burkey. If keith is strolling east from Baron at an average velocity of 3km/hr, and Sue is power-walking west from Burkey at an average velocity of 6km/hr, how long will it take them to meet?
A. 1 minute
B. 2 minutes
C. 3 minutes
D. 6 minutes
Answer:
Time, t = 2 minutes
Explanation:
Given that,
Length of Duplain st. d = 300 m = 0.3 km
If keith is strolling east from Baron at an average velocity of, v = 3 km/hr
Sue is power-walking west from Burkey at an average velocity of, v' = 6 km/hr
To find,
How long will it take them to meet ?
Solution,
When both objects are travelling in opposite direction, then the total speed is given by :
V = v + v'
V = 3 km/hr + 6 km/hr
V = 9 km/hr
Let t is the time taken will it take them to meet. It can be calculated as :
[tex]t=\dfrac{d}{V}[/tex]
[tex]t=\dfrac{0.3\ km}{9\ km/hr}[/tex]
t = 0.033 hour
or
t = 1.98 minutes
i.e. t = 2 minutes
So, they will take 2 minutes to meet. Hence, this is the required solution.
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m². The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.
(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 14.8 rpm. J
(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions
(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in four revolutions. N
Answer:
Part a)
[tex]KE = 101.4 J[/tex]
Part b)
[tex]N = 0.043 revolution[/tex]
Part c)
F = 2.7 N
Explanation:
Part a)
As we know that the rotational kinetic energy of the merry go round is given as
[tex]KE = \frac{1}{2}I\omega^2[/tex]
[tex]KE = \frac{1}{2}84.4(\omega^2)[/tex]
here we know that
[tex]\omega = 2\pi(\frac{14.8}{60})[/tex]
[tex]\omega = 1.55 rad/s[/tex]
Now we have
[tex]KE = \frac{1}{2}(84.4)(1.55^2)[/tex]
[tex]KE = 101.4 J[/tex]
Part b)
Now we know that work done due to torque = change in kinetic energy
[tex]W = KE_f - KE_i[/tex]
[tex]\tau (2N\pi) = 101.4 - 0[/tex]
[tex]375(2\pi N) = 101.4[/tex]
[tex]N = 0.043 revolution[/tex]
Part c)
In order to stop it in four revolutions we have
[tex]\tau(2\pi N) = \Delta KE[/tex]
[tex]FR(2\pi N) = 101.4[/tex]
[tex]F(1.5)(2\pi \times 4) = 101.4[/tex]
F = 2.7 N
The rotational kinetic energy is 100.64 J. The father needs to push approximately 0.0426 revolutions to start the merry-go-round. The required force to stop the merry-go-round after 4 revolutions is 2.68 N.
Explanation:To answer this question, we first need to understand the concept of rotational kinetic energy, which is given by the equation K.E. = 0.5 * I * ω², where I is the moment of inertia and ω is the angular velocity. Next, we must familiarize ourselves with the concept of torque, which relates to the force applied to create rotational motion.
For the rotational kinetic energy when they have an angular velocity of 14.8 rpm, we first convert the angular velocity to rad/s using the formula ω = 2πN/60 where N is in rpm. Plugging in the given values, we get ω = 1.55 rad/s. We then plug I = 84.4 kg·m² and ω = 1.55 rad/s into the kinetic energy equation to get K.E. = 0.5 * 84.4 kg·m² * (1.55 rad/s)² = 100.64 J. For the number of revolutions required to achieve this angular velocity, we first calculate the work done using the formula Work = Torque * Θ, where Θ is the angular displacement in rad. We equate the work done to the kinetic energy we found in the previous part, resulting in 100.64 J = 375 N·m * Θ. Solving for Θ gives us Θ = 0.268 rad. Finally, we convert this to revolutions using the formula 1 rev = 2π rad, yielding 0.0426 revolutions. To calculate the required force to stop the merry-go-round, we use the same Work = Torque * Θ, but this timewe replaceg Θ with the angular displacement for 4 revolutions (4 * 2π rad = 25.12 rad). Equating the work (100.64 J) to the torque gives us the required force F = 100.64 J / (1.50 m * 25.12 rad) = 2.68 N.Learn more about Rotational Motion here:https://brainly.com/question/37888082
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Which type of bonding is found in all molecular substances
Answer:
Covalent bonding
Explanation:
Covalent bonding is the type of bonding found in all molecular substances much as water, carbon dioxide or methane. Unlike ionic bonding which is found in ionic substances, covalent bonding involves sharing, not transfer, of electrons between the bonding atoms to form molecules.
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the acceleration once the sled starts to move
Answer
given,
number of dog = 8
mass of each dog= 19 Kg
mass of sled = 210 Kg
average force = 185 Nss
a) writing all the horizontal force
force acting by dog - friction force = (M + 8m) a
8 F_d - μ m g = (M + 8m) a
assuming coefficient of friction of snow be μ = 0.14
8 x 185 - 0.14 x 210 x 9.8 = (210 + 8 x 19 ) x a
a = 3.29 m/s²
b) the kinetic friction of coefficient is less than static friction
hence, we can suggest that acceleration of the sled will increase once the sled start to move.
a > 3.29 m/s²
What condition is required for cloud formation in the atmosphere?
Answer: To form a cloud, the air that rises must cool to the point where some of the water vapor molecules "clump together" at a faster pace than they are pulled apart by their thermal energy. These molecules then condense to form the clouds and water droplets.
We can see here that Cloud formation in the atmosphere requires two main conditions: the presence of water vapor and the cooling of air.
What is atmosphere?The atmosphere is a layer of gases that surrounds a planet or celestial body, held in place by gravity. On Earth, the atmosphere is the layer of gases that envelops the planet and sustains life. It plays a crucial role in protecting and supporting life on Earth by providing essential elements, regulating temperature, and facilitating various atmospheric processes.
Clouds play a crucial role in the Earth's climate system by reflecting sunlight back into space (resulting in cooling) and trapping heat radiated from the Earth's surface (resulting in warming).
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Refrigerant-134a enters the condenser of a residential heat pump at 800 kPA and 35oC at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power, determine
(a) the COP of the heat pump
(b) the rate of heat absorbtion from the outside air.
Answer:
(A) COP = 2.64
(B) rate of heat absorption= 1.9637 kW
Explanation:
mass flow rate (m) = 0.018 kg/s
work input (Win) = 1.2kW
inlet pressure (P1) = 800kPa
inlet temperature (T1) = 35 degree Celsius
h1 = 271.24 KJ/Kg
outlet pressure (P2) = 800 kPa
outlet temperature (T2) = ?
entalphy (h2) = 95.48 KJ/Kg
The entalphies are gotten from tables for refrigerant 134a at the temperatures and pressures above
(A) COP = Qh ÷ Win
where Qh = m(h1 -h2) from the energy balance equation
Qh = 0.018 ( 271.24 - 95.48 ) = 3.1637 kW
COP = 3.1637 ÷ 1.2 = 2.64
(B) rate of heat absorption = Qh - Win
= 3.1637 - 1.2 = 1.9637 kW
The COP of a heat pump is calculated by using the formula COP = Q_H/W, where Q_H is the heat delivered to the house and W is the work done. Utilizing the specific enthalpies and flow rate, Q_H can be calculated. Subsequently, the rate of heat absorption from the outside can be deduced using Q_L = Q_H - W.
Explanation:The thermal efficiency or COP (Coefficient of Performance) of the heat pump can be calculated using the equation: COP = Q_H/W where Q_H is the heat delivered to the house and W is the work input. In this case, since the fluid leaves the condenser as a saturated liquid, Q_H = m_dot*(h_1 - h_2), where h_1 and h_2 are the specific enthalpies at the state 1 and state 2 and m_dot is the mass flow rate of refrigerant.
From the saturated liquid tables, we know that the specific enthalpy h_2 = 257 kJ/kg for Refrigerant-134a at 800 kPa. Thus, Q_H can be calculated and this value (in kW) can be used to calculate COP.
Then, to answer part (b), the rate of heat absorption from the outside air can be calculated using the equation Q_L = Q_H - W. Q_L is the rate of heat absorption, Q_H is the heat delivered to the house and W is the power absorbed by the compressor.
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A surface receiving sound is moved from its original position to a position three times farther away from the source of the sound. The intensity of the received sound thus becomes A. Nine times higher. B. Nine times lower. C. Three times higher. D. Three times lower.
The intensity of the sound will be B. Nine times lower
Explanation:
The intensity of a sound follows an inverse square law, which means that it is inversely proportional to the square of the distance from the source:
[tex]I\propto \frac{1}{r^2}[/tex]
where
I is the intensity
r is the distance from the source
In this problem, the siund has an intensity of I when the receiver is placed at a distance r from the source.
Later, the receiver is placed three times farther away, so the new distance is
r' = 3r
Therefore, the new intensity of the sound will be:
[tex]I'\propto \frac{1}{r'^2}=\frac{1}{(3r)^2}= \frac{1}{9} (\frac{1}{r^2})= \frac{1}{9}I[/tex]
Therefore, the intensity of the sound received will be nine times lower.
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Answer: B. Nine times lower
Explanation:
An angstrom is about the size of an atom and a fermi is about the size of the nucleus of an atom. Given that 1 angstrom unit = 10⁻¹⁰ m and 1 fermi = 10⁻¹⁵ m, what is the relationship between these units?
Answer:
See explanation below
Explanation:
To do this, we need the relation between a fermi and angstrom. We know the relation between the angstrom and meters, and fermi and meters, so, we can actually solve this by doing the conversion of meters.
1 A = 1x10^-10 m
1 m = 1 A / 1x10^-10 m
1 m = 1x10^10 A
Now if we do the same thing with the fermi:
1 f = 1x10^-15 m
1 m = 1 f / 1x10^-15 m
1 m = 1x10^15 f
then:
1x10^10 A = 1x10^15 f
A/f = 1x10^10 / 1x10^15
A/f = 1x10^-5
A piano tuner hears three beats per second when a tuning fork and a note are sounded together and six beats per second after the string is tightened. What should the tuner do next, tighten or loosen the string? Explain.
Final answer:
The piano tuner should loosen the piano string after hearing an increase from three beats per second to six beats per second upon tightening the string, as this indicates the string's frequency was adjusted away from the tuning fork's frequency.
Explanation:
When a piano tuner hears beats, it indicates that there is a frequency difference between a tuning fork and the piano string that is being compared. Initially, the tuner hears three beats per second, which means the frequency of the piano string is either slightly higher or lower than that of the tuning fork. After tightening the string, the number of beats per second increases to six. This indicates that the frequency of the string has moved further away from the frequency of the tuning fork.
The fact that the beats increased after tightening the string implies that the tuner has adjusted the frequency of the piano string in the wrong direction. Since the goal is to eliminate the beats entirely by matching frequencies, the tuner should loosen the string to reduce the frequency instead of tightening it further.