Mitosis
Penicillium reproduces asexually and forms genetically identical spores by mitosis.
Explanation:
In fungi, spores are produced in a structure called sporangium. This is a bulb-like structure at the tip of the sporangiophore. The parent cells (usually haploid because the fungi spend most of their life cycle in the haploid phase) divides and produces spores by mitosis. This produces haploid spores that are dispersed and readily germinate into haploid hyphae.
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A parallel circuit contains four identical lamps. The current through the energy source is 4 A. The total resistance of the circuit is 10 Ω. What is the voltage across the energy source and across each lamp?
a)2.5 V; 40 V
b)40 V; 40 V
c)2.5 V; 2.5 V
d)40 V; 2.5 V
Answer:
b. 40V , 40V
Explanation:
Connections are as per the figure.
As total current through source is 4A , current through each lamp is 1A.
As total resistance of the circuit is 10Ω ,resistance of each bulb is 40Ω because in case of a parallel circuit in which identical objects are connected , [tex]R_{eff} = \frac{R}{n}[/tex] where R is the resistance of each bulb and n is the number of bulbs.
As per Ohm's law , voltage of the source =IR = 4×10 =40V.
We can see from the figure that if the voltage across the source is 40V , the voltage across each bulb is also 40V.
A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projectile lands on the ground five seconds later. Neglecting air resistance, calculate the projectile’s range and draw a projectile path.
Explanation:
V=40m/s
Vy=V.sina=40.sin20=40 . 0.342=13.68m/s
Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s
Projectile travels during 5 seconds and the ramge becomes:
x=V.t=30.64 . 5=153.2m
Δ→ Δ→
5N 6N
what is the net force?
Answer:
The net force acting on the body, F = 11 N
Explanation:
Given data,
The force acting on the body, f = 5 N
Another force acting on the body, f' = 6 N
The direction of both forces is in the same direction.
The net force is given by the resultant vector,
[tex]F= \sqrt{f^{2} + f'^{2} + 2 ff' Cos \theta}[/tex]
[tex]F= \sqrt{f^{2} + f'^{2} + 2 ff'[/tex] (∵ θ = 0)
[tex]F=\sqrt{(f + f')^{2} }[/tex]
F = f + f'
= 5 N + 6 N
= 11 N
Hence, the net force acting on the body, F = 11 N
you push a book a distance of 5 meters with a force of 10 newtons for 2 seconds how much work did you do on the book
Answer:
50 J
Explanation:
Work, W=Fd where F is the applied force and d is the distance
Substituting 10 N for F and 5 m for d then work done on the book can be expressed as
W=10 N* 5 m=50 Nm= 50 J
Therefore, the work done is equivalent to 50 J
Which element is this?
Answer:Silicon
Explanation:
You add up the protons and neutrons and then look at the element paper
Which structures protect the cell? Check all that apply.
- cell wall
- cell membrane
- mitochondrion
- nucleus
- ribosome
- Golgi body
Answer:
cell wall and cell membrane
Answer:
Cell wall and Cell membrane.
Explanation:
The cell membrane controls what goes in and out of the cell
An electric heater of 1000w is used for 2 hours a day. What is the cost for using in for a month of 28 days, if 1 unit costs 3.00 rupees.
Explanation:
total consumption in a month - 1000×2×28= 56000 w-hrs
1 unit = 1Kwh
total unit in month = 56000/1000 = 56
total cost = 56×3 = 168 rs
12. A volcano erupts on Earth's surface causing a piece of rock with a mass of 95 kg to be ejected from rest to an acceleration of 9.5
m/s2. What force was exerted on the rock by the eruption? (TEK4D)
A 0IN
B 10.0N
C 95 N
D 902.5 N
Answer:
The force exerted on the rock by the eruption is, D. 902.5 N
Explanation:
Given data,
The mass of the rock ejected by the volcano, m = 95 kg
The acceleration of the ejected rock, a 9.5 m/s²
The force acting on an object is defined as the product of the mass and its acceleration. It is given by the relation,
F = m x a
= 95 x 9.5
= 902.5 N
Hence, the force exerted on the rock by the eruption is, F = 902.5 N
A 1800 kg car is parked at the top of a hill 4.7 m high. What is the gravitational potential energy of the car?
Answer:
Gracitational potential energy of the car is 84600J or 84.6KJ
Explanation:
Given that the formula of G.P.E is mgh. So use this formula to find gravitational potential energy :
G.P.E = mgh
m = 1800kg
g = 10m/s
h = 4.7m
G.P.E = 1800×10×4.7
= 84 600J (joules)
= 84.6KJ (kilo joules)
Pls help ASAP this questions are science questions ( biology chemistry physics)
46 I think a
47 I think d
49 I think a
50 I think a
51 idk
54 I think d
55 idk
56 idk
57 idk
58 I think d
59 idk
61 idk
62 I think a
Answer:
It would be A.
Explanation:
The scale goes from 0 to 14, With 0 being acidic and 14 being basic.
So if 7 is neutral, then anything less than 7 is moving more towards being more acidic. Anything higher than 7 is moving towards being more basic.
Answer:
100, 50, 40, 30, 20, 10 butter
Explanation:
I need help with my physics work quickly please!!! I only need help with the circled questions!
Answer:
E11. 0.6 N
E12. 1.61×10⁻⁴ N
E13. 1.12 N
E14. 45 lb
Explanation:
The force due to gravity between two objects is:
F = Gm₁m₂ / r²
where G is the gravitational constant,
m₁ and m₂ are the masses of the objects,
and r is the distance between them.
E11. At a distance r, the force is 9.6 N.
9.6 = Gm₁m₂ / r²
At quadruple the distance (4r), the force is:
F = Gm₁m₂ / (4r)²
F = Gm₁m₂ / (16r²)
F = 9.6 / 16
F = 0.6
E12. Given m₁ = m₂ = 700 kg, and r = 0.45 m:
F = Gm₁m₂ / r²
F = (6.67×10⁻¹¹ m³/kg/s²) (700 kg) (700 kg) / (0.45 m)²
F = 1.61×10⁻⁴ N
E13. At a distance r, the force is 0.28 N.
0.28 = Gm₁m₂ / r²
At half the distance (r/2), the force is:
F = Gm₁m₂ / (r/2)²
F = 4 Gm₁m₂ / r²
F = 4 (0.28)
F = 1.12
E14. The acceleration due to gravity on the Moon is 1/6 that of Earth's. So an astronaut weighs 1/6 as much on the Moon.
W = 270 lb / 6
W = 45 lb
we can see our face clearly on mirror why cant we see our face clear on aluminium or irion plate
I need at least 5 sentence
Because specular reflection occurs with a mirror, while diffuse reflection occurs with aluminium or iron plate
Explanation:
Reflection is a phenomenon typical of waves (such as light wave), that occurs when a wave bounces off the surface of a certain material, going back into the original medium at a certain angle, without changing speed, frequency or wavelength.
The direction of the reflected ray is given by the law of reflection, which states that:
The incoming ray, the reflected ray and the normal to the surface all lie on the same planeThe angle of incidence is equal to the angle of reflectionDepending on the type of surface, two different types of reflection can occur:
Specular reflection: this occurs when the surface is perfectly smooth. In such a case, all the incoming rays of light hit the surface with same angle of incidence, therefore they are reflected at the same angle. This means that a perfect and clear image of the original object can be formed: this is the case of a mirror.Diffuse reflection: when the surface is not smooth, the imperfections on the surface are such that the incoming rays of light hit the surface at different angles of incidence. As a result, the rays are reflected at different angles to each other, and therefore no clear image of the object is produced. This is the case of aluminium or iron plate.Learn more about reflection and other wave phenomena:
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An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the maximum height reached?
Answer:
[tex]\displaystyle y_m=3.65m[/tex]
Explanation:
Motion in The Plane
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of [tex]v_o[/tex] and [tex]\theta\\[/tex] as the initial speed and angle, then we have
[tex]\displaystyle v_x=v_o\ cos\theta[/tex]
[tex]\displaystyle v_y=v_o\ sin\theta-gt[/tex]
[tex]\displaystyle x=v_o\ cos\theta\ t[/tex]
[tex]\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}[/tex]
If we want to know the maximum height reached by the object, we find the value of t when [tex]v_y[/tex] becomes zero, because the object stops going up and starts going down
[tex]\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt[/tex]
Solving for t
[tex]\displaystyle t=\frac{v_o\ sin\theta }{g}[/tex]
Then we replace that value into y, to find the maximum height
[tex]\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2[/tex]
Operating and simplifying
[tex]\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}[/tex]
We have
[tex]\displaystyle v_o=20\ m/s,\ \theta=25^o[/tex]
The maximum height is
[tex]\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}[/tex]
[tex]\displaystyle y_m=3.65m[/tex]
You’ve made the hypothesis that the steeper the slope the faster a ball Will be rolling when it reaches the bottom
What should be your independent variable in an experiment testing this
hypothesis
A. How heavy the Ball is
B. How steep the slope is
C. How high the ball starts on the slope
D. How fast the ball is rolling at the bottom
The neck and bottom of a bottle are 0.5cm and 0.4cm respectively. If the cork in the neck is pressed with a force of 1kgwt and the bottle is full of oil. What is the force in te bottom due to the load
Answer:
0.64 kgwt
Explanation:
You should check that the diameters of the neck and bottom are correct. That is a very small bottle, and the neck shouldn't be bigger than the bottom.
The pressure on the cork = the pressure on the bottom
P₁ = P₂
F₁ / A₁ = F₂ / A₂
F₁ / (π r₁²) = F₂ / (π r₂²)
F₁ / r₁² = F₂ / r₂²
(1 kgwt) / (0.5 cm)² = F / (0.4 cm)²
F = 0.64 kgwt
Round as needed.
A 275-kilogram object starts from rest and accelerates on level ground up to a speed of
41 meters per second over 49.3 meters. It continues at that speed for another 15
meters. At the end of the 15 meters, the object is launched forward at a 37° angle.
a. How much time does it take for the object to reach its final velocity?
b. How much force did the object apply in order to reach that final speed?
c. What are the maximum range and height of the object when it is launched?
d. At what angle would the object have to be launched to reach a height of 50
meters above the ground at the same initial velocity?
e. Describe a realistic scenario in which this might happen.
Answer:
Explanation:
a )
mass m = 275 kg , final velocity v = 41 m/s
distance, s = 49.3 m
acceleration a = ?
v² = u² + 2as
41² = 2 a x 49.3
a = 17.05 m/s²
time t = v - u / a
t = 41 / 17.05
= 2.4 s
b )
force = mass x acceleration
= 275 x 17.05
= 4688.75 N
c ) maximum range = v² / g
= 41²/9.8
= 171.5 m
height ( maximum )
v² sin²45 / 2g ( at angle of projectile of 45 , range and height is maximum.)
= 41²/ 2 x 9.8
= 85.76 m
d )
Let the angle be θ
height h = v² sin²θ / 2g
50 = 41² sin²θ /2g
sin²θ = .58
sinθ = .76
50°.
Please help me people
Answer:
1 ) density = 1200[kg/m^3]
2)
a) wide = 0.35[m]
long = 1,1 [m]
thick = 0.015 [m]
b)
Volume = 0.005775[m^3]
mass = 15.6 [kg]
Explanation:
1)
This problem can be solved by knowing the definition of density which is expressed as the relation between mass and volume, so that the following equation can give us the density value.
[tex]density=\frac{m}{V}\\ where:\\m= mass [kg]\\V= volume [m^3]\\replacing\\density=\frac{90}{0.075} \\density=1200 [kg/m^3][/tex]
2)
The following data is equal to:
a)
wide = 35 [cm] = 0.35[m]
long = 11 [dm] = 11 [decimeters] = 1,1 [m]
thick = 15 [mm] = 0.015 [m]
b)
We know that density is the relationship between mass and volume of a body.
So the volume will be:
volume = wide * long * thick = (0.35*1.1*0.0015) = 0.005775[m^3]
[tex]the mass will be:\\m=density*volume\\m=2700[kg/m^3]*0.005775[m^3]\\m=15.6[kg][/tex]
Answer :
(1) The density of asphalt is, [tex]1200kg/m^3[/tex]
(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.
(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.
Explanation :
Part 1 :
As we are given:
Mass of block = 90 kg
Volume of block = [tex]0.075m^3[/tex]
Formula used :
[tex]\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3[/tex]
Thus, the density of asphalt is, [tex]1200kg/m^3[/tex]
Part 2(a) :
As we are given that:
Length of aluminium sheet = 35 cm
Width of aluminium sheet = 11 dm
Thickness of aluminium sheet = 15 mm
Now we have to convert these dimensions into meters.
Conversions used:
1 cm = 0.01 m
1 dm = 0.1 m
1 mm = 0.001 m
Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m
Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m
Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m
Part 2(b) :
First we have to calculate the volume of aluminium sheet.
Volume of aluminum sheet (cuboid) = Length × Width × Thickness
Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m
Volume of aluminum sheet (cuboid) = 0.005775 m³
Now we have to calculate the mass of aluminium sheet.
[tex]\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}[/tex]
[tex]2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}[/tex]
[tex]\text{Mass of aluminium}=15.59kg[/tex]
Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.
1. a. What is the name of the 2d Century Roman Scientist and Astrologer who created a set of "Handy
Tables of data to compute the position of the Sun, moon and stars?
Answer:
Ptolemy
Explanation:
Answer:
the name of the 2d Century Roman Scientist and Astrologer who created a set of "Handy Tables of data to compute the position of the Sun, moon and stars" is:
Claudius Ptolemy.Explanation:
Ptolemy lived between the year 100 and the year 170 (2nd century), time in which he contributed different bases to astronomy, geography and even astrology. In the first-mentioned area, his geocentric model can be highlighted, with which he mentioned that the Earth It was the center of the universe and therefore it was motionless in the middle of it, and other stars such as the sun or the moon revolved around it and, although in that reasoning it was wrong, it provided fairly accurate means for the time in which the displacement and retrogradation of the planets can be calculated.
If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.) T.F =
-20° C
-30° C
30° C
40° C
Answer:
-20°C
Explanation:
ΔQ = m c Δt
here ΔQ is heat supplied which is 10 calories , m is mass which is 2 g , c is specific heat capacity of ice which is 0.5 cal/(g C°), Δt is change in temperature which here is to be calculated.
re arranging the equation we get
Δt = ΔQ/(mc)
= 10/ ( 2 × 0.5)
= 10/1
= 10 C°
final temperature = -30 +10 = -20°C
hence option is A
I NEED HELP BADLY
While driving her car at 40 m/s, Andrea (with a mass of 50-kg) stops short to avoid hitting a squirrel crossing the road. If the seat belt is in contact with Andrea for 0.5 seconds, determine:
a.) Andrea’s initial momentum
b.) Andrea’s final momentum
c.) What is the change in momentum (impulse)?
d.) What is the force that the seatbelt exerts on Andrea?
Answer:
a) Andrea's initial momentum, 200 kg m/s
b) Andrea's final momentum, 0
c) Impulse, = - 200 Ns
d) The force that the seat belt exerts on Andrea, - 400 N
Explanation:
Given data,
The initial velocity of the car is, u = 40 m/s
The mass of Andrea, m = 50 kg
The time period of deceleration, a = 0.5 s
The final velocity of the car, v = 0
a) Andrea's initial momentum,
p = mu
= 50 x 40
= 200 kg m/s
b) Andrea's final momentum
P = mv
= 50 x 0
= 0 kg m/s
c) Impulse
I = mv - mu
= 0 - 200
= - 200 Ns
The negative sign indicated that the momentum is decreased.
d) The force that the seat belt exerts on Andrea
F = (mv - mu)t
= (0 - 200) / 0.5
= - 400 Ns
Hence,the force that the seat belt exerts on Andrea is, - 400 N
An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).
Answer:
denisity = 52.33 g/c[tex]m^{3}[/tex]
Explanation:
Density:
[tex]d = \frac{m}{v}[/tex]
We have that m = 785 and that v = 15 c[tex]m^{3}[/tex].
[tex]d = \frac{785}{15}[/tex]
d = 52.33 [tex]m^{3}[/tex]
The density of the object, calculated using the formula Density = Mass/Volume with the given mass of 785g and volume of 15cm³, is found to be 52.33 g/cm³.
Explanation:The subject in question is about determining the density of an object, which falls under the area of Physics. In physics, density is defined as mass per unit volume, and the formula for calculating density is Density (ρ) = Mass (m)/Volume (V). In the SI system, we often use grams per cubic centimeter (g/cm³) to express the density of solids and liquids.
To apply the formula to this question, we take the mass of the object, which is given as 785g, and then divide this by the volume of the object, given as 15cm³. Thus, the calculation becomes: ρ = 785g / 15 cm³.
The calculated density, to 2 decimal places, is 52.33 g/cm³. Therefore, the object in question has a density of 52.33 g/cm³.
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I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?
Answer:
The height from which the egg is dropped is 68.54 m.
Explanation:
Given:
Initial velocity of egg is, [tex]u=0\ m/s[/tex](Dropped means initial velocity is 0)
Time taken is, [tex]t=3.74\ s[/tex]
Acceleration experienced by egg is due to gravity, [tex]a=g=9.8\ m/s^2[/tex]
The height from which the egg is dropped is, [tex]d=?[/tex]
Now, we use Newton's equation of motion that relates the distance, initial velocity, time and acceleration.
So, we have the following equation of motion:
[tex]d=ut+\frac{1}{2}at^2[/tex]
Plug in all the given values and solve for 'd'. This gives,
[tex]d=0+\frac{1}{2}\times 9.8\times (3.74)^2\\\\d=\frac{1\times 9.8\times 13.9876}{2}\\\\d=\frac{137.078}{2}\\\\d=68.54\ m[/tex]
Therefore, the height from which the egg is dropped is 68.54 m.
Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:
A) (9.8/1.6)h
B) 1 h
C) the square root of 9.8/1.6 h
D) (1.6/9.8)h
E) the square root of 1.6/9.8 h
A pendulum clock runs at a rate proportional to the inverse square root of the acceleration due to gravity. Since gravity is weaker on the Moon, a clock calibrated for Earth's gravity will run slower on the Moon, taking the square root of 9.8/1.6 hours to complete one revolution, which is answer option C the square root of 9.8/1.6 hours.
The rate at which a pendulum clock runs is inversely proportional to the square root of the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s², while on the Moon, it is about 1.63 m/s². According to the formula for the period of a pendulum (T = 2π √(L/g) where L is the length of the pendulum and g is the acceleration due to gravity), if a clock keeps correct time on Earth, on the Moon it would run at a rate that is the square root of the ratio of the accelerations (the square root of 9.8/1.63). This would mean, for every hour on Earth, the clock on the Moon would show a time of C) the square root of 9.8/1.6 hours.
What is the total displacement of the object ?
Answer:
562.5 m
Explanation:
Displacement is area under the graph of velocity.
d = 10*25 + 25^2/2 = 562.5 m
The displacement of the object is 550 meters.
How to solveThe velocity graph shows that the object is moving in a positive direction for the first 25 seconds.
The velocity is constant at 10 m/s for the first 10 seconds, and then it increases to 20 m/s for the next 15 seconds.
The area under the velocity graph is equal to the displacement of the object. The area of a triangle is equal to (1/2) * base * height. In this case, the base of the triangle is 25 seconds and the height is 20 m/s.
Therefore, the area of the triangle is (1/2) * 25 * 20 = 250 m^2.
The object is also moving in a positive direction for the next 10 seconds, but the velocity is constant at 20 m/s. Therefore, the area under the velocity graph for these 10 seconds is a rectangle with a base of 10 seconds and a height of 20 m/s.
The area of the rectangle is 10 * 20 = 200 m^2.
The total area under the velocity graph is 250 m^2 + 200 m^2 = 550 m^2. Therefore, the displacement of the object is 550 meters.
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What is the acceleration of this object ?
Answer:
-2.5 m/s^2
Explanation:
From a velocity-time graph, the acceleration can be determined by finding the gradient of the straight line graph.
Acceleration=(Change in velocity)/(time)
Where change in velocity = final velocity-initial velocity.
Hence, from the graph,
we can see (10-40)/12 = -2.5
Answer:
-2.5
Explanation:
just did this one
What organism would be at the top of the energy pyramid
A) algae
B) barracuda
C) reef shark
D) snapper fish
A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy for the copper sphere. If the specific heat capacity of copper is 387 J/(kg ⋅ C°), what was the increase in temperature?
0.23 C°
0.81 C°
1.3 C°
2.1 C°
Answer:
Temperature increase = 2.1 [C]
Explanation:
We need to identify the initial data of the problem.
v = velocity of the copper sphere = 40 [m/s]
Cp = heat capacity = 387 [J/kg*C]
The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:
[tex]E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C][/tex]
The correct option is d. 2.1°C. The increase in temperature of the copper sphere was approximately [tex]\( 2.1 \)[/tex]°C, which corresponds to answer choice D).
Given:
- Initial kinetic energy (KE) of the copper sphere = [tex]\( \frac{1}{2} mv^2 \)[/tex]
- Specific heat capacity of copper, [tex]\( c = 387 \)[/tex] J/(kg·°C)
The entire kinetic energy is converted into thermal energy, [tex]\( Q = mc\Delta T \)[/tex], where [tex]\( \Delta T \)[/tex] is the increase in temperature.
1. Calculate the initial kinetic energy:
[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]
2. Set up the equation equating kinetic energy to thermal energy:
[tex]\[ \frac{1}{2} mv^2 = mc\Delta T \][/tex]
3. Solve for [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{\frac{1}{2} mv^2}{mc} \][/tex]
4. Substitute the given values:
[tex]\[ \Delta T = \frac{\frac{1}{2} \times m \times (40)^2}{m \times 387} \][/tex]
5. Simplify the expression:
[tex]\[ \Delta T = \frac{800 \times m}{387 \times m} \]\[ \Delta T = \frac{800}{387} \]\[ \Delta T \approx 2.065 \ \textdegree C} \][/tex]
Rounding to one decimal place, the increase in temperature [tex]\( \Delta T \)[/tex] is approximately [tex]\( 2.1 \)[/tex]°C.
Therefore, the increase in temperature of the copper sphere was approximately [tex]\( 2.1 \)[/tex]°C, which corresponds to answer choice D).
A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy for the copper sphere. If the specific heat capacity of copper is 387 J/(kg ⋅ C°), what was the increase in temperature?
a. 0.23 C°
b. 0.81 C°
c. 1.3 C°
d. 2.1 C°
Is the net force up 0.2 N and down 3 N balanced or not blanced
Answer:
Not balanced
Explanation:
The force pulling the object up is applying less force than the object pulling down, this means the object will move down
Answer:
Not balanced
Explanation:
> Science Sem 2 CR SHS1920
What can happen to solar radiation when it enters Earth's atmosphere? Select three options
It is absorbed by Earth.
It is reflected by clouds.
It is reflected by glacial ice.
It is condensed into visible light.
It is amplified by water molecules.
The following can happen to solar radiation when it enters Earth's atmosphere:
It is absorbed by Earth. It is reflected by clouds. It is reflected by glacial ice.Answer: Option A, B, and C
Explanation:
Solar radiation causes an abnormal effect in humans, if they are exposed directly. All the solar radiation emitted by the sun, does not hit us directly. The energy while reaching us passes through different levels of spheres.
Atmosphere contains nitrogen and oxygen molecules, which absorb some of the solar radiation. And some of it is reflected back by the clouds. And some of the remaining energy is absorbed by the surface of the earth. Likewise, almost 71% of the radiation is absorbed.
Answer: It is absorbed by Earth.
It is reflected by clouds.
It is reflected by glacial ice.
Explanation: Solar radiation is composed of electromagnetic waves that carry energy.
White or shiny surfaces usually reflect some waves (actually almost every surface reflects electromagnetic waves)
So some of the radiation is reflected by clouds or by glacial ice.
The bigest part is actually absorbed by Earth (this is why during the day the temperature increases, a biggest amount of radiation is being absorbed)
What is the name and symbol of the element in the second row and fourteenth column of the periodic table? Hint: Review your periodic table. cobalt (Co) silicon (Si) carbon (C) aluminum (Al)