Answer:
Gamma rays have so much energy they could harm people on Earth. People are protected from gamma rays by Earth's atmosphere. The atmosphere absorbs gamma rays, preventing them from affecting life on Earth. Because gamma rays cannot penetrate Earth's atmosphere, scientists use satellites in space to study them.
Answer:
Explanation:
because gamma rays have so much energy they could harm people on earth
The most successful types of plants on Earth are
Answer:
The angiosperms dominate Earth's surface and vegetation in more environments, particularly terrestrial habitats, than any other group of plants. As a result, angiosperms are the most important ultimate source of food for birds and mammals, including humans.
Explanation:
plz mark brainlest
No me sale este problema :c, plano inclinado
Answer:
i didn't understand,
Explanation:
sorry
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
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.
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.
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
Which of the following statements describes how tectonic plates move?
A. They move from the crust to the core.
B. They move from the mantle to the inner core.
C. They move from the inner core to the outer core.
D. They move slowly on top of the mantle.
Answer:
D
Explanation:
The tectonic plates move on the mantle, sort of floating on it as they are part of the crust. When they collide things like mountain ranges form, and big earthquakes happen.
a ball is launched upward at an angle from the ground. which way does its acceleration point at the top?
For a ball that is launched upward at an angle from the ground, the direction of its acceleration at the top of the projectile curve is downwards.
Acceleration of the ball at the topAs the ball is projected upward its acceleration points upwards until the ball reaches the maximum height or top of the projectile path.
At the top of the projectile path, its acceleration start pointing downwards.
Thus, for a ball that is launched upward at an angle from the ground, the direction of its acceleration at the top of the projectile curve is downwards.
Learn more about downward acceleration here: https://brainly.com/question/22048837
#SPJ1
Answer:
vertical
Explanation:
The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.
Answer:
[tex]W=7.56\times 10^{-19}\ J[/tex]
Explanation:
Given that,
The work function for silver is 4.73 eV.
We need to find the value of the work function from electron volts to joules.
We know that,
[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]
For 4.73 eV,
[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]
So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].
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
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.
An airplane with a ground speed of 550 km/hr travels along a heading of 30 degrees n of w at that high altitude there is a powerful 210 km/hr wind
Answer:
Determine the direction that the pilot is aiming the plane.
Explanation:
(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]
In both the camera and the __________, light enters a narrow opening and is projected onto a photosensitive surface. Group of answer choices
Answer: The HUMAN EYE
Explanation:
The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.
In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.
The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.
The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.
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.
what is the relationship between Hectare and cubic meter
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
1.8kg 42J 9.8 how high is the shelf
Answer:
2.38m
Explanation:
Use potential energy
PE= mgh
42= 1.8*9.8*h
solve for h
to get h= 2.38 m
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.
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
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
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.
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.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]
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
When cars travel around a banked (curved) road at the optimum angle,the normal reaction force (n) can provide the necessary centripetal force without the need for a friction force. (a)Describe whar would happen to Optimum banking angle when radius doubles? (b)Describe what would happen to optimum angle when speed doubles? (c)A car negotiate a turn of radius 80cm.What is the optimum banking angle for this curve if the speed is to be equal to 12m/s?
Answer:
(a) The optimum banking Decreases
(b) The optimum banking Increases
(c) The optimum banking is approximately 86.88°
Explanation:
(a) The equation of motion on a banked road is given as follows;
[tex]v = \sqrt{R \cdot g \times \left(\dfrac{tan (\theta) + \mu}{1 - \mu \cdot tan (\theta) }\right) }[/tex]
For no friction, we have;
v = √(R·g·tan(θ))
Where;
R₁ = The radius of the road
g = The acceleration due to gravity ≈ 9.81 m/s² = Constant
θ₁ = The bank angle
μ = The coefficient pf friction = Constant
v = The vehicle's speed
If the radius doubles, for no friction, we have;
v² = R·g·(tan(θ))
tan(θ) = v²/(R·g)
Therefore, when the radius doubles, tan(θ) becomes smaller and therefore, the optimum banking angle θ decreases (becomes smaller)
(b) When the speed doubles, we have;
v₁ = 2·v
∴ tan(θ₁) = (v₁)²/(R·g) = 4·(v)²/(R·g) = 4·tan(θ)
When the speed doubles, tan(θ) increases and therefore, the optimum banking angle θ increases increases
(c) The radius negotiated by the car, R = 80 cm = 0.8 m
The speed of the car, v = 12 m/s
From tan(θ) = v²/(R·g), we have;
tan(θ) = 12²/(0.8 × 9.81) ≈ 18.349
θ ≈ arctan(18.349°) ≈ 86.88°
A particle is moving on a circular path of radius R. What will be its displacement and distance covered after 3 ½ round?
(please help fast)
Answer:
2r or diameter
Explanation:
After 3 1/2 rounds it will end up on the other side of the circle and displacement will be 2 x the radius = d
Distance = 7 π R
Displacement = 2 R from the starting point directed through the center of the circle
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
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]
What is the relationship between Avogadro's number and a mole?
A. There are 6.02 x 1023 items in a mole, which equals Avogadro's
number
B. A mole is a smaller amount than Avogadro's number.
C. Avogadro's number is used to count particles, and a mole is used
to measure mass.
D. A mole applies to any item, but Avogadro's number is limited to
atoms.
SUBM
Answer:
A. There are 6.02 x 1023 items in a mole, which equals Avogadro's
number
Explanation:
The mole of a substance is
