Answer:
t = 3.44 s
Explanation:
We are given;
Fired from rest, and so; u = 0 m/s
Final speed; v = 170 m/s
Height above flat ground; y_o = 58 m
Height at starting point; y = 0 m
Thus, from Newton's equation of motion, we have;
y - y_o = ut - ½gt²
(since it's motion is against gravity)
Plugging in the relevant values, we have;
0 - 58 = 0 - (½ × 9.8 × t²)
-58 = -4.9t²
t² = 58/4.9
t = √(58/4.9)
t = 3.44 s
What process forms the Mid-Atlantic Ridge?
A. Radioactive decay
B. Seafloor spreading
C. Radiometric dating
D. Sediment formation
Answer:
B. seafloor spreading
Explanation:
divergent motion between the Eurasian and North American, and African and South American Plates. ... In this way, as the plates move further apart new ocean lithosphere is formed at the ridge and the ocean basin gets wider
geological society
& What is meant by expansion effect of heating?
Answer:
it's when heat demolished the object
En la figura, la tensión desarrollada en cada
cuerda está dada por los dinamómetros:
T1=8 N y T2=6 N, y el ángulo de inclinación
de la primera cuerda es de 45°. Determine la
masa de la caja que debe sostener y el
ángulo con respecto a la horizontal.
Answer:
Parte A
El ángulo con respecto al horizonte, de la segunda cuerda es de aproximadamente 19,47°
Parte B
La masa de la caja que se va a sostener es de aproximadamente 0,7808 kg.
Explanation:
Parte A
Los parámetros dados son;
La tensión en la cuerda, T₁ = 8 N
La tensión en la cuerda, T₂ = 6 N
El ángulo de inclinación de la primera cuerda con la horizontal, θ₁ = 45°
Sea θ₂ el ángulo de inclinación de la segunda cuerda, obtenemos;
T₁·cos (θ₁) = T₂·cos (θ₂)
∴ 8 N × cos (45°) = 6 N × cos (θ₂)
cos (θ₂) = 8 N × cos (45°) / (6 N) = (√2)/2 × (4/3) = (2·√2)/3
θ₂ = arcos ((2·√2) / 3) ≈ 19,47°
El ángulo con respecto al horizonte, de la segunda cuerda, θ₂ ≈ 19,47°
Parte B
El peso de la caja, W = T₁·sin (θ₁) + T₂·sin (θ₂)
∴ W = 8 N × sen (45 °) + 6 N × sen (19,47 °) ≈ 7,66 N
El peso de la caja que se va a sostener, W ≈ 7,66 N
La masa de la caja que se va a sostener, m ≈ 7,66 N / (9,81 m/s²) ≈ 0,7808 kg
No me sale este problema :c, plano inclinado
Answer:
i didn't understand,
Explanation:
sorry
In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.
Answer:
1. 36.35 g = 36.35E15 micrometer.
II. 36.35 g = 363.5 millimetre.
III. 36.35 g = 0.03635 kilogram.
Explanation:
Given the following data;
Mass of calorimeter = 36.35 gramsTo convert the mass in grams (g) to;
I. Micrometer
Conversion:
1 g = 1 exp 15 um
36.35 g = X um
Cross-multiplying, we have;
X = 36.35 * 1 exp 15 = 36.35 exp 15 um
36.35 g = 36.35E15 micrometer
II. Millimetre
Conversion:
1 g = 1 milliliter
36.35 g = X milliliter
Cross-multiplying, we have;
X = 36.35 * 1 = 36.35 milliliter
Next, we would convert milliliter to millimetre;
1 milliliter = 10 millimetre
36.35 milliliter = X millimetre
Cross-multiplying, we have;
X = 36.35 * 10 = 363.5 millimetre
36.35 g = 363.5 millimetre
III. Kilogram
Conversion:
1000 grams = 1 kilogram
36.35 g = X kilogram
Cross-multiplying, we have;
X * 1000 = 36.35 * 1
Dividing both sides by 1000, we have;
X = 36.35/1000 = 0.03635 kilogram
36.35 g = 0.03635 kilogram
Note:
g is the symbol for grams.Exp (E) means exponential = 10um is the symbol for micrometer.Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
The softball experiences
force as a result of Amy’s throw. As the ball moves, it experiences
from the air it passes through. It also experiences a downward pull because of
.
Answer:
1.the friction of air, gravity2.gravity
Answer:
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.
Explanation:
Plato
Un alambre de plástico, aislante y recto mide 10 cm de longitud y tiene una densidad de carga de +150 nC/m, distribuidos de manera uniforme por toda su longitud. Se encuentra sobre una mesa horizontal. A) Encuentre la magnitud y la dirección del campo eléctrico que produce este alambre en un punto que está 8 cm directamente arriba de su punto medio. B) Si el alambre ahora se dobla para formar un círculo que se coloca aplanado sobre la mesa, calcule la magnitud y la dirección del campo eléctrico que produce en un punto que se encuentra 6 cm directamente arriba de su centro.
Answer:
English only
Explanation:
When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:
1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ
The solutions:
A) Ey = 4623 N/C
B) Ey = 19.34 N/C
E = Ey = ∫ 2×dE× cosθ
Here cosθ = h/ d ⇒ cosθ = h/√h² + x² dE = K× dQ / d²
d² = h² + x²
k = 8.9 ×10⁹ Nm²C⁻² ; dQ = λ×dx λ = 150×10⁻⁹ C h = 0.08 m
Then by substitution
Ey = 2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²
reordering that equation:
Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³ (2)
To solve the integral we make use of a change of variables
x = h × tanα then x² = h² ×tan²α and dx = h× sec²α dα
plugging that values in equation (2)
Ey = 2×K×λ×h ∫ h× sec²α× dα / [√ ( h² + h²tan²α)]³
Ey = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³ 1 + tan²α = sec²α
Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα
Ey = 2×K×λ/h × ∫ ( 1 / secα dα
Ey = 2×K×λ/h × sinα now we αneed to come back to our original variables:
as x = h × tanα tanα = x/h then x is the opposite leg in a right triangle and h the adjacent one then the hypothenuse is √ (h² + x²) then sin α = x/ √ (h² + x²)
Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵
Ey = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 ) N/C
Ey = 4623 N/C
To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then
Ey = ∫ dE× cosθ
following the same procedure we will find:
Ey = ∫ [K×λ × dl/d²] × h/ d
The importan point here is that the radius of the circle is
2×π×r = 0.01 ( the length of the wire) ⇒ r = 0.16×10⁻² m
And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution
Ey = [K×λ ×h× / ( √ r² + h²)³ ] × 10⁻² N/C
Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)
Ey = 0.8 × 10² / 6
Ey = 19.34 N/C
A cannon sitting on level ground is aimed at 45.0 degrees relative to the horizontal. It fires a test shot at a target located 100.0 meters away from the cannon on the same level ground. The test overshoots the target by 20.0 meters. Which of the following angles can the cannon be adjusted to to hit the target. You may neglect air resistance and assume the cannon always delivers the same initial velocity to the cannonball .
A. 35.9 deg
B. 49.1 deg
C. 28.2 deg
D. 52.8 deg
E. 22.7 deg
Answer:
C. 28.2 deg
Explanation:
The horizontal range of a projectile is given as:
[tex]R = \frac{v^2Sin2\theta}{g}[/tex]
where,
R = Range
v = speed
θ = angle of launch
g = acceleration due to gravity = 9.81 m/s²
First, we will find the launch speed (v) by using the initial conditions:
R = 120 m
θ = 45°
Therefore,
[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]
Now, consider the second scenario to hit the target:
R = 100 m
Therefore,
[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]
Hence, the correct option is:
C. 28.2 deg
what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s
Answer:
[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]
Two drums of the same size and same height are taken.
i)what will be the difference in liquid pressure on their bases if A of them is completely filled and B is half filled and kept at the same place.
ii) what will be the difference in liquid pressure on their bases if both A and B are filled with water completely but one of them is kept at nepal and another in india?why?
iii) what will be the difference in liquid pressure on their bases if A is filled with water and B is filled with salty water and kept at delhi in the same position?why?
Answer:
i) The pressure acting on the base of B will be half the pressure acting on the base of A
ii) The pressure acting on the base of B will be the same as the pressure acting on the base of A
iii) The pressure on the base of drum A will be slightly less than the pressure on the base of drum B
Explanation:
The pressure acting on the base of the drum, P = h·ρ·g
Where;
h = The level of the liquid in the drum
[tex]h_{max}[/tex] = The height of the drums
ρ = The density of the liquid in the drum
g = The acceleration due to gravity ≈ 9.81 m/s²
i) If A is completely filled, we have [tex]h_A[/tex] = [tex]h_{max}[/tex]
Therefore, [tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{liquid}[/tex]×g
If B is half filled, we have, [tex]h_B[/tex] = (1/2)·[tex]h_{max}[/tex]
[tex]P_B[/tex] = (1/2) × [tex]h_{max}[/tex]×[tex]\rho_{liquid}[/tex]×g
Therefore, [tex]P_B[/tex] = (1/2) × [tex]P_A[/tex]
The pressure acting on the base of B will be half the pressure acting on the base of A
ii) If both A and B are each filled with water (the same liquid), then the pressure on their bases will be [tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{water}[/tex]×g = [tex]P_B[/tex], the same, given that the acceleration due to gravity, g, is constant and the same in Nepal and India
iii) If A is filled with water, and B is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;
[tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{water}[/tex]×g < [tex]P_B[/tex] =
The pressure on the base of drum A will be less than the pressure on the base of drum B.
If the radius of curvature of a mirror is 15m and the distance of the object from the mirror is 10m. Find the distance of the image from the mirror and the magnification of the object in meter
Answer:
Data given.
focal length (f)=15m÷2=7.5m
Distance of the object(U)=10m
Image distance (v)=?
Magnification (M)=?
Solution:
From:
1/f=1/u+1/v
1/7.5=1/10+1/v=75
then v=75m
Magnification, M=u/v
=75/10=7.5
Then magnification=7.5
Answer:
v = 30 m and m = 3
Explanation:
Given that,
The radius of curvature of the mirror, R = 15 m
Focal length, f = 7.5 m
Object distance, u = -10 m
We need to find the image distance and the magnification of the object.
Using mirror's formula,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]
The magnification of the object in mirror is given by :
[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]
So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.
(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface is 0.412 m/s2
Answer:
[tex]V.E=498.02m/s^2[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=301Km[/tex]
Gravitational acceleration [tex]g=0.412 m/s^2[/tex]
Generally the equation for Escape velocity is mathematically given by
[tex]V.E^2=2gr[/tex]
[tex]V.E^2=2*0.412m/s^2*301000[/tex]
[tex]V.E^2=248024[/tex]
[tex]V.E=\sqrt{248024}[/tex]
[tex]V.E=498.02m/s^2[/tex]
An ice skater pushes harder with her legs and begins to move faster. Which two laws best describes this
Answer:
Newton' second law and third law describes the situation.
Explanation:
According to the Newton's second law, the force applied on a body is proportional to the rate of change of momentum of the body.
According to the Newton's third law, for every action there is an equal and opposite reaction.
When ice skater pushes harder means more force is applied so he moves fast and more be the action force more be the reaction force.
Thus, Newton' second law and third law describes the situation.
When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre with the strength
Solution :
When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.
The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.
How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Answer:
a. 164°F
b. [tex]91.\overline 1 \ ^{\circ} C[/tex]
c. [tex]140.\overline 4[/tex] kJ
Explanation:
The starting temperature of the water, T₁ = 48F
The temperature at which the water boils, T₂ = 212°F
a. The difference between the initial and the boiling water temperature, ΔT = T₂ - T₁
Therefore;
ΔT = 212°F - 48°F = 164°F
The temperature by which he temperature must be raised, ΔT = 164°F
b. 48°F = ((48 - 32)×5/9)°C = (80/9)°C = [tex]8.\overline 8 \ ^{\circ} C[/tex]
212°F = ((212 - 32)×5/9)°C = 100°C
∴ ΔT = 100°C - [tex]8.\overline 8 \ ^{\circ} C[/tex] = [tex]9.\overline 1 \ ^{\circ} C[/tex]
c. The heat capacity of the water = The heat required to increase four quartz of water by 1 °C = 15.8 kJ
∴ The heat required to raise four quartz of water by [tex]9.\overline 1 \ ^{\circ} C[/tex], ΔQ = 15.8 kJ/°C × [tex]9.\overline 1 \ ^{\circ} C[/tex] = [tex]140.\overline 4[/tex] kJ.
A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000 when hat is slowlh transferred the piston is pushed up through a distance of 4.0cm If 42j of heat is transferred to the system during expansion wht is the change im internal energy of th system
Answer:
10 Joule
Explanation:
The solution and answer are well written in the Pic above.
three condensers are connected in series across a 150 volt supply, the voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10^-8 c.calculate (a) the capacitance of each condenser (b)the effective capacitance of the combination
Answer:
(a) 1.5 nF, 1.2 nF, 1 nF
(b) 0.4 nF
Explanation:
V = 150 V
V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C
(a) C' = q/V' = 6 x 10^-8 / 40 = 1.5 x 10^-9 F
C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F
C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F
(b) The effective capacitance is
[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]
What is matter made of.
Answer:
Matter is made up of atoms
Answer:
Mater is made up of atoms.
Explanation:
Atoms come together to form molecules,which are the building blocks for all types of matter.
If the particles that make up an object begin to move quickly, their average kinetic energy _____ and the object's temperature _____. Group of answer choices
Explanation:
If the particles that make up an object begin to move quickly, their average kinetic energy increases the object's temperature rises. Group of answer choices
A car starts from rest .If its acceleration is 1.5m/s^2 in 1.5 seconds. then calculate the distance traveled by it.
Answer:1.6875 m
Explanation:
Formula= 1/2 x at^2
Is it possible to get a body accelerated even if it is moving with uniform speed? justify.
Answer:Yes, A body can have constant speed but still accelerate as in case of uniform circular motion. In uniform circular motion speed remains constant and direction of velocity changes with every point in the direction of tangent drawn from that point.
Explaination:
Was this helpful
3.00 m^3 of water is at 20.0°C.
If you raise its temperature to
60.0°C, by how much will its
volume expand?
Water
B = 207•10-6 0-1
(Unit = m^3)
Answer:
[tex]\triangle V = 0.02484m^3[/tex]
Explanation:
Given
[tex]V_1 = 3.00m^3[/tex] --- initial volume
[tex]T_1 = 20.0^oC[/tex] --- initial temperature
[tex]T_2 = 60.0^oC[/tex] --- final temperature
[tex]\gamma = 207*10^{-6[/tex] --- coefficient of thermal expansion:
Required
The change in volume
To do this, we make use of cubic expansivity formula
[tex]\triangle V = \gamma * V_2 * (T_2 - T_1)[/tex]
So, we have:
[tex]\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)[/tex]
[tex]\triangle V = 207 * 10^{-6} * 3.00 * 40.0[/tex]
[tex]\triangle V = 0.02484m^3[/tex]
The volume will expand by [tex]0.02484m^3[/tex]
a man is running on the straight road with the uniform velocity of3m/s.calculate its acceleration
Answer:
9m is this the way calculator its acceleration
A man is running on the straight road with the uniform velocity of 3 m/s.
To find out:Acceleration produced by the man.
Solution:Given, the man is running with uniform velocity of 3 m/s.So the velocity did not change.It remains constant.We know, acceleration is the change of velocity per unit time.Since there is no change in velocity in this case, so there will be no acceleration.So, the acceleration produced by the man is zero.Answer:The acceleration produced by the man is zero.Light strikes a smooth wooden tabletop.
What happens to the light after it is reflected?
The light rays bounce off the table and all move in the same direction.
The light rays bounce off the table and move in different directions.
The light rays pass through the table and all move in the same direction.
The light rays pass through the table and move in different directions.
Answer:
For smooth surface:The light rays bounce off the table and all move in the same direction.
A 5.0 kg box moving at 2.0 m/s on a horizontal, frictionless surface runs into a light horizontal spring of force constant 85 N/cm. Use the work-energy theorem to find the maximum compression of the spring.
Answer:
x = 4.85 cm
Explanation:
From work energy theorem when dealing with a spring in compression, we know that total work done is;
W_t = ½kx²
Where;
k is Force constant
x is max compression
Now, we know that this is also equal to the kinetic energy.
K.E = ½mv²
Thus;
½kx² = ½mv²
Making x the subject;
x = √(mv²/k)
We are given;
m = 5 kg
v = 2 m/s
k = 85 N/cm = 8500 N/m
Thus;
x = √(mv²/k)
x = √(5 × 2²/8500)
x = 0.0485 m
x = 4.85 cm
What is this?
Picture
Answer:
may be upside down alphabet :"T"
Explanation:
Which of the following quantity is unit-less? 1 Specific gravity 2 Mass density 3 Acceleration due to gravity 4 All of the above
Answer:
1
Explanation:
Specific gravity is a ratio (of 2 densities) so it has no unit.
why does a desert cooler cool better than a hot dry day
On a hot dry day, the amount of water vapour present in atmosphere is less. Thus, water present inside the desert cooler evaporates more, thereby cooling the surroundings more. Hence, a desert cooler cools better on a hot dry day.
Why are the largest craters we find on the Moon and Mercury so much larger than the largest craters we find on the Earth
Answer:
Because Moon and Mars has no atmosphere.
Explanation:
Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.
When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.
The power of the kettle was 1.5 kW. The 0.2kg heating element took 5 seconds to heat from 20 °C to 100 °C. Calculate the specific heat capacity of water using this information.
Answer:
Specific heat capacity, c = 468.75 J/Kg°C
Explanation:
Given the following data;
Power = 1.5 kW to Watts = 1.5 * 1000 = 1500 Watts
Time = 5 seconds
Mass = 0.2 kg
Initial temperature = 20°C
Final temperature = 100°C
To find specific heat capacity;
First of all, we would have to determine the energy consumption of the kettle;
Energy = power * time
Energy = 1500 * 5
Energy = 7500 Joules
Next, we would calculate the specific heat capacity of water.
Heat capacity is given by the formula;
[tex] Q = mcdt[/tex]
Where;
Q represents the heat capacity or quantity of heat. m represents the mass of an object. c represents the specific heat capacity of water. dt represents the change in temperature.dt = T2 - T1
dt = 100 - 20
dt = 80°C
Making c the subject of formula, we have;
[tex] c = \frac {Q}{mdt} [/tex]
Substituting into the equation, we have;
[tex] c = \frac {7500}{0.2*80} [/tex]
[tex] c = \frac {7500}{16} [/tex]
Specific heat capacity, c = 468.75 J/Kg°C
