Answer:
a) 2·√10 seconds
b) Linda should be approximately 30.6 meters
c) Jenny's speed at the 100-m mark is approximately 6.325 m/s
Explanation:
The speed with which Linda is running = 8.6 m/s
The point Jenny starts = The 80-m mark
The acceleration of Jenny = 1.0 m/s²
a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, t, is given as follows
Δs = u·t + (1/2)·a·t²
Δs = Distance = 100-m - 80-m = 20-m
u = The initial velocity of Jenny = 0
a = Jenny's acceleration = 1.0 m/s²
∴ 20 = 0×t + (1/2) × 1 × t² = t²/2
20 = t²/2
t = √(20 × 2) = 2·√10
The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds
b) The distance Linda runs in t = 2·√10 seconds, d = v × t
Given that Linda's velocity, v = 8.6 m/s, we have;
d = 8.0 × 2·√10 = 16·√10
The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters
Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running
c) Jenny's speed at the 100 m mark is given as follows;
v = u + a·t
t = 2·√10 seconds, a = 1.0 m/s², u = 0
∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325
Jenny's speed at the 100-m mark ≈ 6.325 m/s
The mass of a brick is 2kg. Find the mass of water displaced by it when it is completely immersed in water. (Density of the bricks is 2.5 g/cm^3)
Answer:
2000g
Explanation:
volume=mass/density
=2000/2.5
=800cm³
mass=density×volume
=800×2.5
=2000g
22) How is it possible to fill medicine in a syringe?explain
I hope this helps you ^-^
when is the mass of an object if it exerts a force of 160 N and an acceleration of 8.15m/s^2
Answer:
f=ma.......m=f/a......m=20kg
A light spring with force constant 3.05 N/m is compressed by 7.80 cm as it is held between a 0.400-kg block on the left and a 0.800-kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push the blocks apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is 0, 0.035, and 0.397. (Let the coordinate system be positive to the right and negative to the left. Indicate the direction with the sign of your answer. Assume that the coefficient of static friction is the same as the coefficient of kinetic friction. If the block does not move, enter 0.)
Complete Question
A light spring with force constant 3.05 N/m is compressed by 7.80 cm as it is held between a 0.400-kg block on the left and a 0.800-kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push the blocks apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is 0, 0.035, and 0.397. (Let the coordinate system be positive to the right and negative to the left. Indicate the direction with the sign of your answer. Assume that the coefficient of static friction is the same as the coefficient of kinetic friction. If the block does not move, enter 0.)
(a) u = 0 heavier block m/s2 m/s2 lighter block
(b)M = 0.035 heavier block m/s2 m/s2 lighter block
Answer:
a) [tex]A_h=0.297[/tex]
[tex]A_l=-0.59475[/tex]
b) [tex]a=0[/tex]
[tex]a=-0.25175m/s^2[/tex]
Explanation:
From the question we are told that:
Force constant [tex]k=3.05N/m[/tex]
Compression Length [tex]l_c=7.80cm=0.07m[/tex]
Left Mass [tex]M_l=0.400kg[/tex]
Right Mass [tex]M_r=0.800kg[/tex]
Coefficient of kinetic friction [tex]\mu=0, 0.035, and\ 0.397.[/tex]
Therefore
Spring force is given as
[tex]F_s=Kx[/tex]
[tex]F_s=3.05*0.070[/tex]
[tex]F_s=0.238N[/tex]
Generally the equation for Acceleration is mathematically given by
[tex]A=\frac{F}{m}[/tex]
For Heavier block
[tex]A_h=\frac{F_s}{m_r}[/tex]
[tex]A_h=\frac{0.238N}{0.8}[/tex]
[tex]A_h=0.297[/tex]
For Lighter blocks
[tex]A_l=\frac{F_s}{m_r}[/tex]
[tex]A_l=\frac{-0.238N}{0.4}[/tex]
[tex]A_l=-0.59475[/tex]
b)
Generally the equation for Force is mathematically given by
[tex]F_s-F=ma[/tex]
For Heavier block
[tex]F>Fs[/tex]
Therefore
[tex]a=0[/tex]
For Lighter blocks
[tex]F-F_s=ma[/tex]
[tex](0.035)(0.4)(9.8)-(0.2379)=(0.4)a[/tex]
[tex]a=-0.25175m/s^2[/tex]
The gravitational force acting on various masses is measured on different planets. Measured values for the forces acting on the corresponding masses are shown in the data table. Analyze the data and develop a method for comparing the gravitational field strengths on the different planets. Use your method to compare the gravitational field strengths, and report your conclusions.
who is a meteorologist
Answer:
A meteorologist is a scientist that studies and examines Earth's atmosphere and its interactions with Earth's oceans, land, and biosphere.
Explanation:
Meteorologists use math and science to understand and predict changes in weather and climate.
The boiling point of water is 1000 C at sea level. The boiling point of butane is -1.50C… If we leave liquid butane in a bowl on a table in a room where the temperature is 240C, butane will
A. evaporate.
B. condense.
C. freeze.
D. melt.
Answer: If we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.
Explanation:
A temperature at which the the liquid and gaseous phase of a substance of a substance are present in equilibrium with each other is called boiling point.
For example, the boiling point of butane is -1.5 degree Celsius.
This means that at a temperature above -1.5 degree Celsius, butane will exist is gaseous state. That is, at a temperature of 24 degree Celsius butane will evaporate.
Thus, we can conclude that if we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.
PLEASE HELP!!!
Write the sentences in your copybook and draw a line through one of the words in
bold to complete each of these sentences about alkali metals correctly.
Alkali metals generally become more / less dense going down the group.
The melting and boiling points of alkali metals increase / decrease down the group.
The softness of alkali metals increases / decreases going down the group.
The speed with which alkali metals react with oxygen increases / decreases going
down the group.
Answer:
Densities increase down the group
MP and BP decrease down the group
Softness increased going down the group
Speed of reacting increases going down the group
A policeman kicks in a door with a force of 4500 N. What force does the door apply to the policeman’s leg?
Answer:
-4500 N
Source: Brainly
The police officer must be angry 0_0
Please help me with this...
And write all steps..
Answer:
[tex]2\frac{m}{s^2} =a[/tex]Explanation:
Use the kinematic equation.
[tex]v_{2} =v_{1} +at[/tex]This equation can be derived from [tex]f=ma[/tex], but we can just memorize, or look them up when needed as it saves us time.
Now we can plug our measurements into each variable to solve for acceleration.
[tex]18\frac{m}{s} =8\frac{m}{s} +a*5s[/tex]Subtract 8m/s from both sides.
[tex]10\frac{m}{s} =a*5s[/tex]Divide by 5 seconds. Left with acceleration in terms of [tex]\frac{m}{s^2}[/tex]
[tex]2\frac{m}{s^2} =a[/tex]. Una varilla de cobre de coeficiente de dilatación 1,4*10-5 °C -1 , tiene una longitud de 1.20 metros a una temperatura ambiente de 18 ˚C . ¿Cuál sera su longitud 100 ˚C
Answer:
La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.
Explanation:
Asumiendo que la varilla de cobre experimenta deformaciones muy pequeñas y que las deformaciones no longitudinales son despreciables con respecto a las deformaciones longitudinales, la deformación longitudinal de la varilla se estima mediante la siguiente fórmula:
[tex]l_{f} = l_{o}\cdot [1+\alpha \cdot (T_{f}-T_{o})][/tex] (1)
Donde:
[tex]l_{o}[/tex] - Longitud inicial de la varilla, en metros.
[tex]\alpha[/tex] - Coeficiente de dilatación, en [tex]^{\circ}C^{-1}[/tex].
[tex]T_{o}[/tex] - Temperatura inicial de la varilla, en grados Celsius.
[tex]T_{f}[/tex] - Temperatura final de la varilla, en grados Celsius.
Si sabemos que [tex]l_{o} = 1.20\,m[/tex], [tex]\alpha = 1.4\times 10^{-5}\,^{\circ}C^{-1}[/tex], [tex]T_{o} = 18\,^{\circ}C[/tex] y [tex]T_{f} = 100\,^{\circ}C[/tex], entonces la longitud final de la varilla es:
[tex]l_{f} = (1.20\,m)\cdot \left[1 + \left(1.4\times 10^{-5}\,^{\circ}C^{-1}\right)\cdot (100\,^{\circ}C-18\,^{\circ}C)\right][/tex]
[tex]l_{f} = 1.201\,m[/tex]
La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.
An airplane starts from rest and undergoes a uniform acceleration of 8.1 m/s2 for 19.4 s seconds before leaving the ground. What is its displacement?
Answer:
GIVEN:
v₀=0ms⁻¹
a= 8.1ms⁻²
t= 19.4s
REQUIRE:
d=?
CALCULATUION:
as we know,
d=v₀t+1/2at²
by putting values
d=0ms⁻¹×19.4s+1/2×8.1ms⁻²×(19.4s)²
d=0m+1/2×8.1ms⁻²×376.36s²
d=1/2×3048.516m
d=1524.258m
d≈1524m
. When 2 moles of helium gas expand at constant pressure P = 1.0 × 105 Pascals, the temperature increases from 2℃ to 112℃. If the initial volume of the gas was 45 liters. Cp=20.8J/mol.K, Cv=12.6J/mol.K. Determine i. The work done W by the gas as it expands (4) ii. The total heat applied to the gas (2) iii. The change in internal energy (2
Answer:
i. Work done by the gas as it expands is approximately 1,900 J
ii. The total heat supplied is approximately 4, 576 J
iii. The change in internal energy is approximately 2,772 J
Explanation:
The constant pressure of the helium gas, P = 1.0 × 10⁵ Pa
The initial and final pressure of the gas, T₁, and T₂ = 2°C (275.15 K) and 112°C (385.15 K) respectively
The number of moles of helium in the sample of helium gas, n = 2 moles
The volume occupied by the gas at state 1, V₁ = 45 L
i. By ideal gas law, we have;
P·V = n·R·T
Therefore;
[tex]V = \dfrac{n \cdot R \cdot T}{P}[/tex]
Plugging in the values gives;
[tex]V_2 = \dfrac{n \cdot R \cdot T_2}{P}[/tex]
Where;
V₂ = The volume of the gas at state 2
Therefore;
[tex]V_2 = \dfrac{2 \cdot 8.314 \cdot 385.15}{1.0 \times 10^5} \approx 0.064[/tex]
The volume of the gas at state 2, V₂ ≈ 0.064 m³ = 64 Liters
Work done by the gas as it expands, W = P × (V₂ - V₁)
∴ W ≈ 1.0 × 10⁵ Pa × (64 L - 45 L) = 1,900 J
Work done by the gas as it expands, W ≈ 1,900 J
ii. The total heat supplied, Q = Cp·n·ΔT
∴ Q = 20.8 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 4,576 J
The total heat supplied, Q = 4, 576 J
iii. The change in internal energy, ΔU = Cv·n·ΔT
∴ ΔU = 12.6 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 2,772 J
The change in internal energy, ΔU = 2,772 J
Which unit is equivalent to J/s?
O A. Meters
O B. Watts
O C. Newtons
O D. Calories
B. Watts
Then j/s is the rate of transferring energy or doing work. Its unit is the Watt, equivalent to 1 Joule per second.
HELPPPPPPPPPPP PLEASEEEEEEEEEEE
Complete this sentence. The solubility of a sample will ____________ when the size of the sample increases.
stay the same
decrease
increase
be unable to be determined
the answer is not decrease
The solubility of the sample will decrease
Which labels are correct for the regions marked? a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium b. X: Faster in gases than liquids Y: Slowest in solids Z: Faster in liquids than gases c. X: Slower in solids than liquids Y: Velocity depends on medium Z: Faster in liquids than gases d. X: Velocity depends on medium Y: Fastest in gases Z: Slower in liquids than solids
Answer:
a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium.
Explanation:
Speed of sound is fastest in solids. Sound waves travel more quickly in solid, than of liquid and gases. Sound waves travel most slowest in gases. Speed of sound varies significantly and it depends upon medium it is travelling through. In more rigid medium sounds velocity will be faster.
What is the energy of an electromagnetic wave that has a frequency of
4.0 x 109 Hz? Use the equation E = hf, where h = 6.626 x 10-34 J·s.
Answer:
[tex] Energy, \; E = 2.6504 * 10^{-34} \; Joules [/tex]
Explanation:
Given the following data;
Frequency = 4.0 x 10⁹ Hz
Planck's constant, h = 6.626 x 10-34 J·s.
To find the energy of the electromagnetic wave;
Mathematically, the energy of an electromagnetic wave is given by the formula;
E = hf
Where;
E is the energy possessed by a wave.
h represents Planck's constant.
f is the frequency of a wave.
Substituting the values into the formula, we have;
[tex] Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34} [/tex]
[tex] Energy, \; E = 2.6504 * 10^{-34} \; Joules [/tex]
please answer quick for brainlist ; )
Answer:
The diagram assigned B
explanation:
Check the direction of the two vectors, their resultant must be in the same direction.
an image of a statue appears to be 11.5cm behind convex mirror with focal length 13.5cm. Find the distance from the statue to the mirror
Answer:
77.625 cm
Explanation:
The given distance of the image behind the convex mirror, v = 11.5 cm
The focal length of the mirror, f = 13.5 cm
The mirror formula for convex mirror is given as follows;
[tex]\dfrac{1}{u} - \dfrac{1}{v} = -\dfrac{1}{f}[/tex]
Where;
u = The distance from the statue to the mirror
Therefore, we get;
[tex]\dfrac{1}{u} = -\dfrac{1}{f} + \dfrac{1}{v}[/tex]
Plugging in the values gives;
[tex]\dfrac{1}{u} = -\dfrac{1}{13.5} + \dfrac{1}{11.5} = \dfrac{8}{621}[/tex]
∴ The distance from the statue to the mirror, u = 621/8 cm = 77.625 cm.
A distressed car is rolling backward, downhill at 3.0 m/s when its driver finally manages to
get the engine started. What velocity will the car have 6.0 s later if it can accelerate at
3.0 m/s??
Answer:
Explanation:
Acceleration is equal to the change in velocity over the change in time, or
[tex]a=\frac{v_f-v_i}{t}[/tex] where the change in velocity is final velocity minus initial velocity. Filling in:
[tex]3.0=\frac{v_f-(-3.0)}{6.0}[/tex] Note that I made the backward velocity negative so the forward velocity in our answer will be positive.
Simplifying that gives us:
[tex]3.0=\frac{v_f+3.0}{6.0}[/tex] and then isolating the final velocity, our unknown:
3.0(6.0) = v + 3.0 and
3.0(6.0) - 3.0 = v and
18 - 3.0 = v so
15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.
A basketball is shot by a player at a height of 2.0m. The initial angle was 53° above the horizontal. At the highest point, the ball was travelling 6 m/s. If he scored (the ball went through the rim that is 3.00m above the ground), what was the player's horizontal distance from the basket?
At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that
v cos(53°) = 6 m/s ==> v = (6 m/s) sec(53°) ≈ 9.97 m/s
The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are
x = (9.97 m/s) cos(53°) t = (6 m/s) t
y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²
Find the times t for which the ball reaches a height of 3.00 m:
3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²
==> t ≈ 0.137 s or t ≈ 1.49 s
The second time is the one we care about, because it's the one for which the ball would be falling into the basket.
Now find the distance x traveled by the ball after this time:
x = (6 m/s) (1.49 s) ≈ 8.93 m
Which sentence best describes a role of gravity in the formation of the
universe?
A. Gravity caused the universe to expand from a central point.
B. Gravity caused background microwave radiation to be emitted as
the universe formed.
C. Gravity caused galaxies to move apart from one another in a
symmetrical way.
D. Gravity caused stars to come together and galaxies to form after
the big bang
Answer:
I think it's option D
Explanation:
I think it's option D but not so sure
Which image illustrates refraction?
Answer:
B illustrates refraction
Even through there is equal and opposite reaction,usually the two forces are not seen balanced.Why?
Answer:
This may refer to a situation like:
"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"
Remember the second Newton's law:
F = m*a
suppose that the mass of the person is 3 times the mass of the box.
So, if the box has a mass M, the person will have a mass 3*M
Then the Newton's equation for the box when the person pushes with a force F is:
F = M*a
solving for the acceleration, we get:
F/M = a
While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:
F = (3*M)*a'
Solving for the acceleration, we get:
F/(3M) = a'
Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).
Is easy to see that the acceleration of the box is 3 times the acceleration of the person.
So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.
This is why in situations like this, the forces do not seem balanced.
Encuentre la presion en la otra seccion estrecha si las velocidades en las secciones son de 0.50m\sy 2m\s
Answer:
ΔP = 1875 Pa, P₂ = P₁ - 1875
Explanation:
Let's use Bernoulli's equation, with the subscript 1 for the widest Mars and the subscript 2 for the narrowest part, suppose that the pipe is horizontal
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P₁ -P₂ = ½ ρ (v₂² - v₁²)
suppose the fluid is water
P₁ - P₂ = ½ 1000 (2² - 0.5²)
ΔP = 1875 Pa
this is the pressure difference between the two sections
the pressure in the narrowest section is
P₂ = P₁ - 1875
After landing the aeroplane momentum becomes zero .Explain how the law of conservation helds here.
Answer:
see the explanation below
Explanation:
Momentum is a product of the mass of a particle and its velocity.
and also, momentum is a vector quantity; i.e. it has both magnitude and direction.
Now a plane in the air has both magnitude and velocity
When the plane lands the velocity will amount to zero although the mass is still very much intact
Now the mass* zero velocity= zero
Hence when a plane lands the momentum is zero
A bus Starts from rest. If the acceleration of bus become 10 m/s2 after 15 sec Calculate the final Velocity of the bus
what change will occur in gravitational force between two bodies if mass of both object is doubled and distance between their center is halved
The gravitational force between 2 bodys decreases with distance between the two bodies.
f=G m1m2/r2
Answer:
if the distance between 2 objects is halved than the gravitation doubles ,as gravitation is inversely propotional, between the distance of 2 objects.Which of the following are true about simile and metaphor? Select all that apply.
similes and metaphors make comparisons for emotional effect
simile uses "like" or "as": metaphor does not
similes and metaphors make comparisons to give us mental pictures
both are the same
Answer:
It is simile uses "like" or "as": metaphor does not
Explanation:
it just it
Answer:
third one
Explanation:
i cant think of anymore
What's the resultant of the 3 forces?
Answer:
Explanation:
We need to find the x-components of each of these vectors and then add them together, then we need to find the y-components of these vectors and then add them together. Let's get to that point first. That's hard enough for step 1, dontcha think?
The x-components are found by multiplying the magnitude of the vectors by the cosine of their respective angles, while the y components are found by multiplying the magnitude of the vectors by the sine of their respective angles.
Let's do the x-components for all the vectors first, so we get the x-component of the resultant vector:
[tex]F_{1x}=12 cos0[/tex] and
[tex]F_{1x}=12[/tex]
[tex]F_{2x}=9cos90[/tex] and
[tex]F_{2x}=0[/tex]
[tex]F_{3x}=15 cos126.87[/tex] and
[tex]F_{3x}=-9.0[/tex] (the angle of 126.87 is found by subtracting the 53.13 from 180, since angles are to be measured from the positive axis in a counterclockwise fashion).
That means that the x-component of the resultant vector, R, is 3.0
Now for the y-components:
[tex]F_{1y}=12sin0[/tex] and
[tex]F_{1y}=0[/tex]
[tex]F_{2y}=9sin90[/tex] and
[tex]F_{2y}=9[/tex]
[tex]F_{3y}=15sin126.87[/tex] and
[tex]F_{3y}=12[/tex]
That means that the y-component of the resultant vector, R, is 21.
Put them together in this way to find the resultant magnitude:
[tex]R_{mag}=\sqrt{(3.0)^2+(21)^2}[/tex] which gives us
[tex]R_{mag}=21[/tex] and now for the angle. Since both the x and y components of the resultant vector are positive, our angle will be where the x and y values are both positive in the x/y coordinate plane, which is Q1.
The angle, then:
[tex]tan^{-1}(\frac{21}{3.0})=82[/tex] degrees, and since we are QI, we do not add anything to this angle to maintain its accuracy.
To sum up: The resultant vector has a magnitude of 21 N at 82°
