Answer:
The total heat required to melt the ice is approximately 3.473 MJ
Explanation:
The given parameters for the layer of ice are;
The thickness of the layer of ice, t = 0.80 cm = 0.008 m
The area of the wind shield, A = 1.4 m²
The initial temperature of the ice, T₁ = -2.0 °C = 271.15 K
The density of the ice, ρ = 917 kg/m
The temperature at which ice melts, T₂ = 0 °C = 273.15 K
We have;
The mass of the ice, m = ρ × t × A
∴ m = 917 kg/m³ × 0.008 m × 1.4 m² = 10.2704 kg
The specific heat capacity of ice, c = 2,090 J/(kg·K)
∴ The equation for the heat capacity of the ice to melt, is given as follows;
ΔQ = m·c·ΔT
Where;
ΔT = T₂ - T₁
∴ ΔT = T₂ - T₁ = 273.15 K - 271.15 K = 2 K
ΔQ = 10.2704 kg × 2,090 J/(kg·K) × 2 K = 42.930272 kJ
The latent heat to melt the ice, Q = The latent heat of fusion of ice, L × Mass of ice, m
The latent heat of fusion of ice, L = 334 kJ/kg
∴ Q = 334 kJ/kg × 10.2704 kg = 3,430.3136 kJ
The total heat required to melt the ice, H = ΔQ + Q
∴ H = 42.930272 kJ + 3,430.3136 kJ = 3,473.243872 kJ ≈ 3.473 MJ.
A 6 kg object's Ug increases by 150 J. What was its change in height?
Please help I don’t understand this and fast please
Answer:
2.5 m
Explanation:
Potential energy is the energy stored in an object as a result of its position relative to other objects
The change in potential energy is given by:
ΔPE = mgh;
where ΔPE is the change in potential energy, m is the mass if the object, g is the acceleration due to gravity and h is the change in height of the object.
Hence given that g = 10 m/s², ΔPE = 150 J, m = 6 kg, hence:
ΔPE = mgh
150 = 6 * 10 * h
150 = 60h
h = 2.5 m
Hence the change in height is 2.5 m
If the final velocity is 0. third equation of motion will be
Answer:
vf²=vi²+2a∆x
Explanation:
The third equation of motion gives the final velocity of an object under uniform acceleration given the distance traveled and an initial velocity: v 2 = v 0 2 + 2 a d . v^2=v_0^2+2ad. v2=v02+2ad. The graph of the motion of the object.
What is an effect of continental drift?
Answer: An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.
How does 'g' vary from place to place?
Explanation:
The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.
A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.5 kg person stands on the platform at a distance of 1.05 m from the center, and a 28.3 kg dog sits on the platform near the person 1.43 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
I_syst = 278.41477 kg.m²
Explanation:
Mass of platform; m1 = 117 kg
Radius; r = 1.61 m
Moment of inertia here is;
I1 = m1•r²/2
I1 = 117 × 1.61²/2
I1 = 151.63785 kg.m²
Mass of person; m2 = 62.5 kg
Distance of person from centre; r = 1.05 m
Moment of inertia here is;
I2 = m2•r²
I2 = 62.5 × 1.05²
I2 = 68.90625 kg.m²
Mass of dog; m3 = 28.3 kg
Distance of Dog from centre; r = 1.43 m
I3 = 28.3 × 1.43²
I3 = 57.87067 kg.m²
Thus,moment of inertia of the system;
I_syst = I1 + I2 + I3
I_syst = 151.63785 + 68.90625 + 57.87067
I_syst = 278.41477 kg.m²
maize is a monocotyledonous seed and pea is a dicotyledonous seed why? give short and the suitable answer I will mark you as a brainelist
Answer:
A dicot is a flowering plant that has one seed leaves. The monocot plants have a single cotyledon. Maize only has one cotyledon in their seed, so it's a monocot. Seeds having two Cotyles are mainly called a Dicot. A pea is a dicotyledonous plant, the seed (the pea itself) has two halves, cotyledons, hence dicot being 2.
Explanation:
One or more of the cotyledons are the first to appear from a germinating seed. Based on the number of cotyledons, botanists classify flowering plants (angiosperms) into :
a) plants with one embryonic leaf, termed monocotyledonous (monocots).
b) plants with two embryonic leaves, termed dicotyledonous (dicots).
Helpful Link:
https://www.vedantu.com/question-answer/in-pea-caster-and-maize-the-number-of-cotyledons-class-11-biology-cbse-5f626a17e5bde9062ff6d2a3
Define electric current and drift velocity.
Answer:
Explanation: The voltage or potential difference between two points is defined to be the change in ... I = qnAv relates the drift velocity to the current
determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3
Answer:
divide the density of solution by density of water
EXPLANATION:
LIKE:
1.25÷1000kgm-3
Water falls down as a stream from a tap. Why does it not scatter?
Answer:
as it hits the ground it's dispersed and this causes it to move in different directions
matter can enter and exit which of the following systems?
A. isolated only
B. open only
C. both open and isolated
D. Both closed and isolated
A car travels first 10 km in 20 minutes and another 10 km in 30 minutes. What is the average speed of the car in m/s?
Total distance = 10 km + 10 km = 20 km
1 km = 1000 m
20km x 1000 = 20,000 m
Total time = 20 min. + 30 min. = 50 minutes
Average speed = Distance / time
Average speed = 20,000/50 min
Average speed = 400 m/s
state the term used to describe the turning force exerted by the man pushing down on a lever to lift one end of a heavy log
The term used to describe the turning force exerted when pushing an object is effort.
What is effort?The is the input force or force applied at one point of a lever in order to overcome a load.
The relationship between effort, load and mechanical advantage of a lever is given as;
M.A = L/E
where;
M.A is mechanical advantageL is the loadE is the applied force or effortThus, the term used to describe the turning force exerted when pushing an object is effort.
Learn more about effort here: https://brainly.com/question/24237657
A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied by the club is 115 N. What is the speed of the ball as it leaves the tee
Answer:
v = 50.5 m/s
Explanation:
F = (m)(^v/^t)
115N = (0.04551kg)(v/(0.020s))
2,526.917161 m/s² = v/(0.020s)
v = 50.53834322 m/s
v = 50.5 m/s
Please help! Will mark Brainliest.
Answer:
18 Nm
Explanation:
if the correct answer
Derive the following equations. :a=u-v by t
1.
"An estimate of the future value of some variable" is the definition of
A. procurement
B. logistics.
C. a forecast.
D. capacity
Which of the following is a vector quantity? i. Force ii. Velocity iii. Acceleration iv. All of these 5771
Option ( iv ) is the correct answer.
☛ DefinitionA vector quantity the physical quantity that has both direction as well as magnitude.
Observe the given figure and find the the gravitational force between m1 and m2.
Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
[tex]F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}[/tex]
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
[tex]F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N[/tex]
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
explain how force and surface area affect the applied pressure.
Answer:
force and surface area are two factors affecting pressure on solids
more the force you apply, more will be the pressure
pressure and force are directly proportional meaning if Force is greater, pressure will also be greater
more the surface area of the solid less will be the pressure
surface area and pressure are inversely proportional meaning if surface are is big, pressure will be less, surface area small, pressure will be greater
Answer and Explanation:
We have a basic equation: Pressure = Force/Area.
So for example:
Increase pressure - increase the force or reduce the area the force acts on.
Decrease pressure - decrease the force, or increase the area the force acts on
The force per unit area is pressure. The force on the object is spread over the surface area. The area where the force is applied is divided by the equation for pressure.
Cheers,
; (b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.
Explanation:
(b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.
A low-power laser used in a physics lab might have a power of 0.50 mW and a beam diameter of 3.0 mm. Calculate:a.The average light intensity of the laser beam.b. The intensity of a lightbulb producing 100-W light viewed from 2.0 m.c.Compare the intensity of the laser to the intensity of the lightbulb. Is it advisable to look directly at a laser
Answer:
A) I_laser = 70.74 W/m²
B) I_bulb = 1.989 W/m²
C) it is not advisable to look at the laser beam directly.
Explanation:
We are given;
Power; P = 0.50 mW = 0.5 × 10^(-3) W
Diameter; d = 3 mm = 0.003 m
Radius; r = d/2 = 0.003/2 = 0.0015 m
A) Area of beam; A = πr²
A = 0.0015²π
Now, formula for average intensity is;
I = P/A
I = (0.5 × 10^(-3))/0.0015²π
I = 70.74 W/m²
B) We are told to find the intensity of a lightbulb producing 100-W.
Thus, P = 100 W
A light bulb is spherical in shape. Thus;
Area; A = 4πr²
We are told it's 2 m away.
Thus; r = 2 m
A = 4π(2)²
A = 16π
Thus, I = P/A = 100/16π
I = 1.989 W/m²
C) The intensity of the laser beam is far greater than that of the light bulb. Thus, it is not advisable to look at the laser beam directly.
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. You reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 \mathrm { m } / \mathrm { s } ^ { 2 }10m/s 2 . a. How much distance is between you and the deer when you come to a stop
Answer:
Explanation:
Discount the time here; it's not important. It doesn't tell you how long it takes the car to stop, it only refers to reaction time, which means nothing in the scheme of things.
The useful info is as follows:
initial velocity = 20 m/s
final velocity = 0 m/s
a = -10 m/s/s
and we are looking for the displacement. Use the following equation:
[tex]v^2=v_0^2+2a[/tex]Δx
where v is the final velocity, v₀ is the initial velocity, a is the deceleration (since it's negative), and Δx is displacement. Filling in:
[tex]0^2=(20)^2+2(-10)[/tex]Δx and
0 = 400 - 20Δx and
-400 = -20Δx so
Δ = 20 meters
The refractive index of water is 1.33
and that of diamond is 2.42. Draw a labelled diagram to show how a light ray bends when it travels from water
into diamond.
Explanation:
light travel slower in daimond
A 5.41 kg ball is attached to the top of a vertical pole with a 2.37 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle θ, between 0∘ and 90∘, that the string makes with the pole. Use g=9.81 m/s2.
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g
(1 -cos²)/ cos θ = [tex]\frac{v^2 }{g \ L}[/tex]
1 - cos² θ = [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex] cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]
T cos θ = m g
resolved
tan θ = [tex]\frac{v^2}{ r g}[/tex]
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
What is the purpose of the lab the importance of the topic and the question you are trying to answer?
perpose of lab is to store apparatus and do some experiment
Helps someone to know the exert lengh of something
Write a balanced nuclear reaction for one complete cycle
Answer:
H + H ------> He + energy. He + He -----> H + H + He.
Explanation:
Hydrogen having one proton and no neutron fuse with hydrogen having one proton and one neutron forming helium atom with the release of photon. After that two helium atoms combine together forming two hydrogen atoms having one proton each whereas one helium atom having two protons and two neutrons present in their nucleus so the end product of this reaction is hydrogen atoms and helium.
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.25 ms2.25 ms from an initial speed of 4.25 m/s4.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.75 kg1.75 kg
Answer:
The force is 3305.6 N.
Explanation:
Final velocity, v = 0
time, t = 2.25 ms
initial velocity, u = 4.25 m/s
mass, m = 1.75 kg
Let the acceleration is a.
Use first equation of motion.
v = u + a t
0 = 4.25 + a x 0.00225
a = - 1888.9 m/s^2
The force is
F = ma
F = 1.75 x 1888.9
F = 3305.6 N
Phương trình chuyển động thẳng đều của một chất điểm có dạng: x = 2t – 10. (x: km, t: h). Quãng đường đi được của chất điểm sau 2h là bao nhiêu?
Answer:
Distance cover in 2 hour = 6 kilometer
Explanation:
Given equation:
x = 2t - 10
where
x = kilometer
t = hour
Find:
Distance cover in 2 hour
Computation:
T = 2
So,
x = 2t - 10
x = 2(2) - 10
x = 4 - 10
x = -6
Distance cover in 2 hour = 6 kilometer
Answer:
The distance is 6 km.
Explanation:
The equation of uniform linear motion of a particle has the form: x = 2t – 10. (x: km, t: h). What is the distance traveled by the particle after 2 hours?
x = 2t - 10
distance traveled after t = 2 hours
Substitute t = 2 in the given expression
x = 2 x 2 - 10
x = 4 - 10
x = - 6 km
So, the distance is 6 km.
A solenoid passing by a current of 5.0 A generates a magnetic field at its diameter of 50 μT. Thus the number of spirals per length scale is:
A. 5.0 / π Spear / m
B. 10 / π Spear / m
C. 20 / π Spear / m
D. 25 / π Spear / m
Answer:
D. 25 / π Spiral / m
Explanation:
Given;
current, I = 5 A
magnetic field strength, B = 50 μT = 50 x 10⁻⁶ T
The magnetic field strength is given as;
[tex]B = \mu_0 nI\\\\where;\\\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi \times 10^{-7} T/A.m\\\\n \ is \ the \ number \ of \ spirals \ per \ length\\\\n = \frac{B}{\mu_0 I} = \frac{50 \times 10^{-6}}{5\times 4\pi \times 10^{-7}} = \frac{25}{\pi } \ spirals /m \\\\[/tex]
Therefore, the correct option is D. 25 / π Spiral / m
A spring has a spring constant of 5 N/m and is stretched 10 m. What its U,?
a. 50 J
b. 70 J
c. 90 J
d. 250 J
Answer
I am not sure but it is may be 50J
I hope that's, this answer is fine.
