Answer:
mass=56.12kg
Explanation:
PE=mgh
4620=m×9.8×8.4
make m subject of the formula...
m =4620/(9.8×8.4)
m=4620/82.32
m=56.12kg
calculate the amount of energy needed to take 34g of ice at -2°C to 118°C
Answer:
Q = 8608.8 J
Explanation:
Given that,
The mass of ice, m = 34 g
The temperature changes from -2°C to 118°C.
The specific heat of ice, c = 2.11 J/g°C
The heat energy needed,
[tex]Q=mc\delata T\\\\Q=34\times 2.11\times (118-(-2))\\Q=8608.8\ J[/tex]
So, 8608.8 J of energy is needed.
URGENT!!! PLEASE HELP
Answer:
guard cell
hope it helps
"" damped force equation ""
Answer:
In many vibrating systems the frictional force Ff can be modeled as being proportional to the velocity v of the object: Ff = −cv, where c is called the viscous damping coefficient.
Una muestra de agua (líquida) de 1220 kg se encuentra a 0° C y baja se temperatura hasta -29°C mientras se congela en el proceso. ¿Cuánta energía es liberada al ambiente (Mega Joules) ?
Answer:
-148,6 MJ
Explanation:
Dado que la liberación de calor al medio ambiente es;
H = mcθ
Dónde;
m = masa de agua
c = capacidad calorífica específica del agua
θ = aumento de temperatura
H = 1220 kg × 4200 × [-29-0]
H = -148,6 MJ
How do you use the periodic table to recall the ionic charge of an alkali metal, an alkaline earth metal, or aluminum?
The positive charge is the group number.
The negative charge is the group number.
The positive charge is the period number.
The negative charge is the period number.
Answer:
the positive charge is the period number
Explanation:
I might be wrong
Answer:
The positive charge is the group number.
Explanation:
A point charge of +3.0 X 10-7 coulomb is placed 2.0 X 10-2 meter from a second point charge of +4.0 X 10-7 coulomb. What is the magnitude of the electrostatic force on the charges?
A. 6.0 X 10-12 N
B. 3.0 X 10-10 N
C. 5.4 X 10-2 N
D. 2.7 N
Answer:
D. 2.7 N
Explanation:
Applying
F = kq'q/r²................ Equation 1
Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge
From the question,
Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 1
F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²
F = 26.94×10⁻¹ N
F = 2.694 N
F ≈ 2.7 N
Which of the following statements is incorrect?
1) A ray of light travels from a point A to point B in a path that takes the shortest time
2) In reflection angle of incidence equal to angle of reflection
3) Magnification for a plane mirror is always unity
4) A plane mirror can form only virtual images
Answer:
2
Explanation:
in reflection the incident ray and reflected rays are equal an angle
20 POINTS!!! The photograph shows a the result of waves made by an earthquake.
What was transported to this location by the earthquake waves?
A. Gravity
O B. Energy
C. Rock
D. Matter
Answer:
It's B = Energy ....
i hope its right
Question 1 of 35
Which statement applies only to magnetic force instead of both electric and
magnetic forces?
A. It can push objects apart.
B. It acts between a north pole and a south pole.
C. It acts between objects that do not touch.
D. It can pull objects together.
How are ions formed ?
Answer:
Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the ..
Answer:
Ions are formed when atoms lose or gain electrons, simply when electrons from the metal transfers electrons from its outermost shell( forming a positively charged metallic ion) to the outermost shell of the non-metallic atom( forming a negatively charged ion)
If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car (Assume the bottom of the cliff is zero)
Group of answer choices
the gravitational potential energy is decreasing
the gravitational potential energy has not changed
the gravitational potential energy is increasing
Explanation:
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.
so the answer is "the gravitational potential energy is decreasing"
velocity of B rays ?
Answer:
is from 9 x 107 m/sec to 27 x 107 m/s
HELP ASAP
Which list places different units of matter in the correct sequence from
largest to smallest?
A. Buildings, bricks, rock particles, protons, atoms
B. Buildings, rock particles, bricks, atoms, protons
C. Buildings, bricks, atoms, rock particles, protons
D. Buildings, bricks, rock particles, atoms, protons
Answer:
I think it's D!!
cuz protons are in the atoms
Where is the water table located?
Answer:
The water table is the upper surface of the zone of saturation. The zone of saturation is where the pores and fractures of the ground are saturated with water. It can also be simply explained as, the upper level, below which the ground is saturated.
Brains or Brawl? List the reason why you choose what you choose.
What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What is the direction of this force?
Answer:
88.34 N directed towards the center of the circle
Explanation:
Applying,
F = mv²/r................... Equation 1
F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.
But,
v = 2πr/t................... Equation 2
Where t = time, π = pie
Substitute equation 2 into equation 1
F = m(2πr/t)²/r
F = 4π²r²m/t²r
F = 4π²rm/t²............. Equation 3
From the question,
Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s
Constant: π = 3.14
Substitute these values into equation 3
F = 4(3.14²)(0.7)(0.8)/0.5²
F = 88.34 N directed towards the center of the circle
What are the poles of a bar magnet.is it North West South or east.
Answer:
One end of any bar magnet will always want to point north if it is freely suspended. This is called the north-seeking pole of the magnet, or simply the north pole. The opposite end is called the south pole.
Explanation:
Answer:
North Pole and South Poles
Explanation:
There is no east or west in a magnetic field of a magnet. The magnetic field lines comes from North pole to south pole.
An object dropped from a cliff falls with a constant acceleration of 10 m/s2.Find its speed 5 s after it was dropped.
Answer:
50m/s2
Explanation:
u=0m/s
a=10m/s
v=?
t=5s
v=u+at
=0+(10×5)
=0+50
=50m/s
Evaluate (5.7x106 kg) * (6.3x10-2 m/s2) and express the answer in
scientific notation.
Answer:
Force = 3.591 * 10⁵ Newton
Explanation:
Given the following data;
Mass = 5.7 * 10⁶ kg
Acceleration = 6.3 * 10^-2 m/s²
To express the force as a scientific notation;
Force = mass * acceleration
Force = 5.7 * 10⁶ * 6.3 * 10^-2
Force = 359100 Newton
In scientific notation;
A number is said to be written in scientific notation when we multiply the numbers between 1 and 10 by a power of 10.
Force = 3.591 * 10⁵ Newton
A 15 cm length of wire is moving perpendicularly
through a magnetic field of strength 1.4 T at the rate
of 0.12 m/s. What is the EMF induced in the wire?
A. OV
C. 0.025 v
B. 0.018 V
D. 2.5 V
Answer: C or B
Explanation:
The EMF induced in the wire moving perpendicularly through a magnetic field is 0.025V. The correct option is C.
What is EMF?The EMF is the electro motive force which causes the current to induce in the object moving in the magnetic field.
Given is the length of wire L =15cm =0.15m, magnetic field strength B = 1.4T and velocity of wire V =0.12 m/s
EMF is related to the length of wire, magnetic field strength and velocity of wire proportionally.
ε = B x L x V
Plug the values, we get
ε = 1.4 x 0.15 x 0.12
ε = 0.025 Volts
Thus, the correct option is C.
Learn more about EMF.
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the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament is a blackbody radiator? The filament of a particular electric lamp can be considered as a 90%blackbody radiator. calculate the energy per second radiated when its temperature is 2000k if its surface area is 10∧-6 m²
Answer:
(a) [tex]\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}[/tex]
(b) P = 0.816 Watt
Explanation:
(a)
The power radiated from a black body is given by Stefan Boltzman Law:
[tex]P = \sigma AT^4[/tex]
where,
P = Energy Radiated per Second = ?
σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴
T = Absolute Temperature
So the ratio of power at 250 K to the power at 2000 K is given as:
[tex]\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}[/tex]
(b)
Now, for 90% radiator blackbody at 2000 K:
[tex]P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4[/tex]
P = 0.816 Watt
A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.0 m/s without any friction. When a car is going 31.8 m/s on this curve, what minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?
Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of
a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²
There are two forces acting on the car in this situation:
• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n
• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)
By Newton's second law, the net forces in the perpendicular and parallel directions are
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) = ma
==> sin(θ) = a/g ==> θ = arcsin(a/g) ≈ 17.8°
(Notice that in the paralell case, the positive direction points toward the center of the curve.)
When rounding the curve at 31.8 m/s, the car's radial acceleration changes to
a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²
and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) + f = ma
The first equation gives
n = mg cos(θ)
and substituting into the second equation, we get
mg sin(θ) + µmg cos(θ) = ma
==> µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12
Answer:
Explanation:
You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is
[tex]F_c=\frac{mv^2}{r}[/tex] where
[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,
m is the mass of the car,
v is the velocity with which the car is moving, and
r is the radius of the circle that the car is moving around.
For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become
[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by
f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).
Going on and getting buried even deeper,
[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:
μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:
μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:
μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is
μ = 1.4
The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,
On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.
Wind, moving water, sunlight, and heat from earth's interior are sources of
which principle is used in mass spectrograph to estimate the mass of a charged particle
M^3 is a derived unit why?
Answer:
From the base unit of length, we can define volume, and from the base units of length, mass, and time, we can define energy. ... Volume. Since volume is length cubed, its SI derived unit is m3
Explanation:
Hope it helps
Hai tiếp điểm hình trụ bằng đồng có đầu hình cầu bán kính R = 70mm, lực ép tiếp điểm F = 100N, Mô đun đàn hồi của đồng là 11.8 * 106(N/cm2). Cho biết: a = 0.86 * (F*R/E)1/3, điện trở suất của đồng là 1.62 * 10-8 (Ω/m), với tiếp điểm đồng thì hệ số K1 = 3,16.10-4 (ΩN1/2). Biết các giá trị ứng suất của đồng là: σdh = 38300 N/cm2; σdẻo = 45000 N/cm2; σd.Nát = 51000 N/cm2. Tính điện trở tiếp xúc trong trường hợp thỏa mãn điều kiện tản dòng lí tưởng và trong trường hợp thực nghiệm.
Answer:
5dhydydyeze6ese6z7zuezuze6ezyezyezzeze
What is the KE of a bird, mass 40g flying at 23m/s?
Answer:
230
Explanation:
You do Ke = kinetic times distance divide by mass
Question 3 of 5
Which term refers to a transfer of thermal energy between objects?
O A. Temperature
OB. Heat
O C. Kinetic energy
O D. Potential energy
Answer:
Heat
Explanation:
The term heat refers to the conduction process which occurs when energy is passed between objects
Answer
Heat
did quiz and got it correct
A fixed mass of gas has a volume of gas of 25cm3. the pressure of the gas is 100kPA. the volume of the gas is slowly decreased by 15cm3 at a constant temperature. what is the change in the pressure of the gas?
a) 67kPA
b) 150kPA
c) 170kPA
d) 250kPA
give reasons
A fixed mass of gas has a volume of 25 [tex]cm^3[/tex], the pressure of the gas is 100 kPa, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.
What is the calculation of the change in pressure?PV = nRT (P= pressure of the gas, V =volume, n = number of moles of gas, R = gas constant, and T =temperature of the gas in kelvin)
Suppose the gas is an ideal gas and that the temperature is constant,
P1V1 = P2V2
Here P1 = 100 kPa, V1 = 25 [tex]cm^3[/tex], V2 = 10 [tex]cm^3[/tex],
100 kPa x 25 [tex]cm^3[/tex] = P2 x 10 [tex]cm^3[/tex]
P2 = (100 kPa x 25 [tex]cm^3[/tex]) / 10 [tex]cm^3[/tex]
P2 = 250 kPa
the change in pressure of the gas is,
ΔP = P2 - P1 = 250 kPa - 100 kPa = 150 kPa
The reason is that when the volume of a fixed mass of gas is decreased, the pressure of the gas increases proportionally, so here assuming that the temperature is constant it is calculated.
Hence, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.
Learn more about the calculation of the change in pressure here.
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State Pascal’s law. (2) b) The area of one end of a U-tube is 0.01 m2 and that of the other
end the force 1 m2, when a force was applied on the liquid at the first end, the force experienced at
the other end was
Answer:
Pascal Law's says that:
If the area of one end of a U-tube is A, and the area of the other end is A'. then if we apply a force F in the first end (the one of area A), the force experienced at the other end must be:
F' = F*(A'/A).
b) Now we can apply this to our particular case:
if the area of one end is 0.01m^2, and the area of the other end is 1m^2
Then we have:
A = 0.01m^2
A' = 1m^2
So, if now we apply a force F in the first end, the force experienced at the other end will be:
F' = F*(1m^2/0.01m^2) = F*100
This means that the force in the other end must be 100 times the force in the first end.