Answer:
18 Nm
Explanation:
if the correct answer
PLEASE HELP!!!
Write the sentences in your copybook and draw a line through one of the words in
bold to complete each of these sentences about alkali metals correctly.
Alkali metals generally become more / less dense going down the group.
The melting and boiling points of alkali metals increase / decrease down the group.
The softness of alkali metals increases / decreases going down the group.
The speed with which alkali metals react with oxygen increases / decreases going
down the group.
Answer:
Densities increase down the group
MP and BP decrease down the group
Softness increased going down the group
Speed of reacting increases going down the group
A policeman kicks in a door with a force of 4500 N. What force does the door apply to the policeman’s leg?
Answer:
-4500 N
Source: Brainly
The police officer must be angry 0_0
Please help me with this...
And write all steps..
Answer:
[tex]2\frac{m}{s^2} =a[/tex]Explanation:
Use the kinematic equation.
[tex]v_{2} =v_{1} +at[/tex]This equation can be derived from [tex]f=ma[/tex], but we can just memorize, or look them up when needed as it saves us time.
Now we can plug our measurements into each variable to solve for acceleration.
[tex]18\frac{m}{s} =8\frac{m}{s} +a*5s[/tex]Subtract 8m/s from both sides.
[tex]10\frac{m}{s} =a*5s[/tex]Divide by 5 seconds. Left with acceleration in terms of [tex]\frac{m}{s^2}[/tex]
[tex]2\frac{m}{s^2} =a[/tex]A solution has a pH of 8. Which best describes the solution?
a strong acid
a strong base
a weak acid
a weak base
Answer:
D
Explanation:
a weak base is the answer
Answer:
weak base because its near to 7 (neutral) and above 7 is base
What is the S.I Unit of electric flux?
Answer:
Volt metre (Vm)
or Newton/Coulomb x (Metre)^2 (N/C x m^2)
Question 8 of 16
A book with a mass of 1.8 kg sits on a bookshelf. If it has a gravitational
potential energy of 42 J, how high is the shelf? The acceleration of gravity is
9.8 m/s2
A. 2.1 m
B. 0.9 m
C. 1.6 m
D. 2.4 m
SUBMIT
Answer:
Explanation:
PE = mgh where m is the mass, g is gravity, and h is the height from which the object can potentially fall. We are given enough info here to solve for h. Filling in:
42 = 1.8(9.8)h and
[tex]h=\frac{42}{(1.8)(9.8)}[/tex] and rounding to 2 sig fig's as we need:
h = 2.4 m, choice D
. Una varilla de cobre de coeficiente de dilatación 1,4*10-5 °C -1 , tiene una longitud de 1.20 metros a una temperatura ambiente de 18 ˚C . ¿Cuál sera su longitud 100 ˚C
Answer:
La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.
Explanation:
Asumiendo que la varilla de cobre experimenta deformaciones muy pequeñas y que las deformaciones no longitudinales son despreciables con respecto a las deformaciones longitudinales, la deformación longitudinal de la varilla se estima mediante la siguiente fórmula:
[tex]l_{f} = l_{o}\cdot [1+\alpha \cdot (T_{f}-T_{o})][/tex] (1)
Donde:
[tex]l_{o}[/tex] - Longitud inicial de la varilla, en metros.
[tex]\alpha[/tex] - Coeficiente de dilatación, en [tex]^{\circ}C^{-1}[/tex].
[tex]T_{o}[/tex] - Temperatura inicial de la varilla, en grados Celsius.
[tex]T_{f}[/tex] - Temperatura final de la varilla, en grados Celsius.
Si sabemos que [tex]l_{o} = 1.20\,m[/tex], [tex]\alpha = 1.4\times 10^{-5}\,^{\circ}C^{-1}[/tex], [tex]T_{o} = 18\,^{\circ}C[/tex] y [tex]T_{f} = 100\,^{\circ}C[/tex], entonces la longitud final de la varilla es:
[tex]l_{f} = (1.20\,m)\cdot \left[1 + \left(1.4\times 10^{-5}\,^{\circ}C^{-1}\right)\cdot (100\,^{\circ}C-18\,^{\circ}C)\right][/tex]
[tex]l_{f} = 1.201\,m[/tex]
La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.
An airplane starts from rest and undergoes a uniform acceleration of 8.1 m/s2 for 19.4 s seconds before leaving the ground. What is its displacement?
Answer:
GIVEN:
v₀=0ms⁻¹
a= 8.1ms⁻²
t= 19.4s
REQUIRE:
d=?
CALCULATUION:
as we know,
d=v₀t+1/2at²
by putting values
d=0ms⁻¹×19.4s+1/2×8.1ms⁻²×(19.4s)²
d=0m+1/2×8.1ms⁻²×376.36s²
d=1/2×3048.516m
d=1524.258m
d≈1524m
. When 2 moles of helium gas expand at constant pressure P = 1.0 × 105 Pascals, the temperature increases from 2℃ to 112℃. If the initial volume of the gas was 45 liters. Cp=20.8J/mol.K, Cv=12.6J/mol.K. Determine i. The work done W by the gas as it expands (4) ii. The total heat applied to the gas (2) iii. The change in internal energy (2
Answer:
i. Work done by the gas as it expands is approximately 1,900 J
ii. The total heat supplied is approximately 4, 576 J
iii. The change in internal energy is approximately 2,772 J
Explanation:
The constant pressure of the helium gas, P = 1.0 × 10⁵ Pa
The initial and final pressure of the gas, T₁, and T₂ = 2°C (275.15 K) and 112°C (385.15 K) respectively
The number of moles of helium in the sample of helium gas, n = 2 moles
The volume occupied by the gas at state 1, V₁ = 45 L
i. By ideal gas law, we have;
P·V = n·R·T
Therefore;
[tex]V = \dfrac{n \cdot R \cdot T}{P}[/tex]
Plugging in the values gives;
[tex]V_2 = \dfrac{n \cdot R \cdot T_2}{P}[/tex]
Where;
V₂ = The volume of the gas at state 2
Therefore;
[tex]V_2 = \dfrac{2 \cdot 8.314 \cdot 385.15}{1.0 \times 10^5} \approx 0.064[/tex]
The volume of the gas at state 2, V₂ ≈ 0.064 m³ = 64 Liters
Work done by the gas as it expands, W = P × (V₂ - V₁)
∴ W ≈ 1.0 × 10⁵ Pa × (64 L - 45 L) = 1,900 J
Work done by the gas as it expands, W ≈ 1,900 J
ii. The total heat supplied, Q = Cp·n·ΔT
∴ Q = 20.8 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 4,576 J
The total heat supplied, Q = 4, 576 J
iii. The change in internal energy, ΔU = Cv·n·ΔT
∴ ΔU = 12.6 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 2,772 J
The change in internal energy, ΔU = 2,772 J
Which unit is equivalent to J/s?
O A. Meters
O B. Watts
O C. Newtons
O D. Calories
B. Watts
Then j/s is the rate of transferring energy or doing work. Its unit is the Watt, equivalent to 1 Joule per second.
HELPPPPPPPPPPP PLEASEEEEEEEEEEE
Complete this sentence. The solubility of a sample will ____________ when the size of the sample increases.
stay the same
decrease
increase
be unable to be determined
the answer is not decrease
The solubility of the sample will decrease
Which labels are correct for the regions marked? a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium b. X: Faster in gases than liquids Y: Slowest in solids Z: Faster in liquids than gases c. X: Slower in solids than liquids Y: Velocity depends on medium Z: Faster in liquids than gases d. X: Velocity depends on medium Y: Fastest in gases Z: Slower in liquids than solids
Answer:
a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium.
Explanation:
Speed of sound is fastest in solids. Sound waves travel more quickly in solid, than of liquid and gases. Sound waves travel most slowest in gases. Speed of sound varies significantly and it depends upon medium it is travelling through. In more rigid medium sounds velocity will be faster.
What is the energy of an electromagnetic wave that has a frequency of
4.0 x 109 Hz? Use the equation E = hf, where h = 6.626 x 10-34 J·s.
Answer:
[tex] Energy, \; E = 2.6504 * 10^{-34} \; Joules [/tex]
Explanation:
Given the following data;
Frequency = 4.0 x 10⁹ Hz
Planck's constant, h = 6.626 x 10-34 J·s.
To find the energy of the electromagnetic wave;
Mathematically, the energy of an electromagnetic wave is given by the formula;
E = hf
Where;
E is the energy possessed by a wave.
h represents Planck's constant.
f is the frequency of a wave.
Substituting the values into the formula, we have;
[tex] Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34} [/tex]
[tex] Energy, \; E = 2.6504 * 10^{-34} \; Joules [/tex]
please answer quick for brainlist ; )
Answer:
The diagram assigned B
explanation:
Check the direction of the two vectors, their resultant must be in the same direction.
an image of a statue appears to be 11.5cm behind convex mirror with focal length 13.5cm. Find the distance from the statue to the mirror
Answer:
77.625 cm
Explanation:
The given distance of the image behind the convex mirror, v = 11.5 cm
The focal length of the mirror, f = 13.5 cm
The mirror formula for convex mirror is given as follows;
[tex]\dfrac{1}{u} - \dfrac{1}{v} = -\dfrac{1}{f}[/tex]
Where;
u = The distance from the statue to the mirror
Therefore, we get;
[tex]\dfrac{1}{u} = -\dfrac{1}{f} + \dfrac{1}{v}[/tex]
Plugging in the values gives;
[tex]\dfrac{1}{u} = -\dfrac{1}{13.5} + \dfrac{1}{11.5} = \dfrac{8}{621}[/tex]
∴ The distance from the statue to the mirror, u = 621/8 cm = 77.625 cm.
A distressed car is rolling backward, downhill at 3.0 m/s when its driver finally manages to
get the engine started. What velocity will the car have 6.0 s later if it can accelerate at
3.0 m/s??
Answer:
Explanation:
Acceleration is equal to the change in velocity over the change in time, or
[tex]a=\frac{v_f-v_i}{t}[/tex] where the change in velocity is final velocity minus initial velocity. Filling in:
[tex]3.0=\frac{v_f-(-3.0)}{6.0}[/tex] Note that I made the backward velocity negative so the forward velocity in our answer will be positive.
Simplifying that gives us:
[tex]3.0=\frac{v_f+3.0}{6.0}[/tex] and then isolating the final velocity, our unknown:
3.0(6.0) = v + 3.0 and
3.0(6.0) - 3.0 = v and
18 - 3.0 = v so
15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.
A basketball is shot by a player at a height of 2.0m. The initial angle was 53° above the horizontal. At the highest point, the ball was travelling 6 m/s. If he scored (the ball went through the rim that is 3.00m above the ground), what was the player's horizontal distance from the basket?
At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that
v cos(53°) = 6 m/s ==> v = (6 m/s) sec(53°) ≈ 9.97 m/s
The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are
x = (9.97 m/s) cos(53°) t = (6 m/s) t
y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²
Find the times t for which the ball reaches a height of 3.00 m:
3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²
==> t ≈ 0.137 s or t ≈ 1.49 s
The second time is the one we care about, because it's the one for which the ball would be falling into the basket.
Now find the distance x traveled by the ball after this time:
x = (6 m/s) (1.49 s) ≈ 8.93 m
Which sentence best describes a role of gravity in the formation of the
universe?
A. Gravity caused the universe to expand from a central point.
B. Gravity caused background microwave radiation to be emitted as
the universe formed.
C. Gravity caused galaxies to move apart from one another in a
symmetrical way.
D. Gravity caused stars to come together and galaxies to form after
the big bang
Answer:
I think it's option D
Explanation:
I think it's option D but not so sure
Which image illustrates refraction?
Answer:
B illustrates refraction
Even through there is equal and opposite reaction,usually the two forces are not seen balanced.Why?
Answer:
This may refer to a situation like:
"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"
Remember the second Newton's law:
F = m*a
suppose that the mass of the person is 3 times the mass of the box.
So, if the box has a mass M, the person will have a mass 3*M
Then the Newton's equation for the box when the person pushes with a force F is:
F = M*a
solving for the acceleration, we get:
F/M = a
While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:
F = (3*M)*a'
Solving for the acceleration, we get:
F/(3M) = a'
Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).
Is easy to see that the acceleration of the box is 3 times the acceleration of the person.
So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.
This is why in situations like this, the forces do not seem balanced.
Encuentre la presion en la otra seccion estrecha si las velocidades en las secciones son de 0.50m\sy 2m\s
Answer:
ΔP = 1875 Pa, P₂ = P₁ - 1875
Explanation:
Let's use Bernoulli's equation, with the subscript 1 for the widest Mars and the subscript 2 for the narrowest part, suppose that the pipe is horizontal
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P₁ -P₂ = ½ ρ (v₂² - v₁²)
suppose the fluid is water
P₁ - P₂ = ½ 1000 (2² - 0.5²)
ΔP = 1875 Pa
this is the pressure difference between the two sections
the pressure in the narrowest section is
P₂ = P₁ - 1875
After landing the aeroplane momentum becomes zero .Explain how the law of conservation helds here.
Answer:
see the explanation below
Explanation:
Momentum is a product of the mass of a particle and its velocity.
and also, momentum is a vector quantity; i.e. it has both magnitude and direction.
Now a plane in the air has both magnitude and velocity
When the plane lands the velocity will amount to zero although the mass is still very much intact
Now the mass* zero velocity= zero
Hence when a plane lands the momentum is zero
An upright image which reduced in size 10 times occurred in a mirror. If the radius of
curvature of the mirror is 2 m, bow far is the object from the mirror?
How to solve?
Answer:
p = -9 m
Explanation:
For this exercise we use the equation of the geometric optics constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively.
The mirrors the focal length is
f = R / 2
f = 2/2
f = 1 m
the magnification is
m =[tex]\frac{h'}{h} = - \frac{q}{p}[/tex]
indicates that the image was reduced h ’= h/10 implies that m = 1/10
[tex]\frac{1}{10} = - \frac{q}{p}[/tex]
we write our system of equations
p = -10q
1/1 = [tex]\frac{1}{p} + \frac{1}{q}[/tex]
we substitute
1 = [tex]\frac{1}{p} - \frac{10}{p}[/tex]
1 = 1/p (1 - 10)
1 = -9 / p
p = -9 m
A bus Starts from rest. If the acceleration of bus become 10 m/s2 after 15 sec Calculate the final Velocity of the bus
what change will occur in gravitational force between two bodies if mass of both object is doubled and distance between their center is halved
The gravitational force between 2 bodys decreases with distance between the two bodies.
f=G m1m2/r2
Answer:
if the distance between 2 objects is halved than the gravitation doubles ,as gravitation is inversely propotional, between the distance of 2 objects.Which of the following are true about simile and metaphor? Select all that apply.
similes and metaphors make comparisons for emotional effect
simile uses "like" or "as": metaphor does not
similes and metaphors make comparisons to give us mental pictures
both are the same
Answer:
It is simile uses "like" or "as": metaphor does not
Explanation:
it just it
Answer:
third one
Explanation:
i cant think of anymore
What's the resultant of the 3 forces?
Answer:
Explanation:
We need to find the x-components of each of these vectors and then add them together, then we need to find the y-components of these vectors and then add them together. Let's get to that point first. That's hard enough for step 1, dontcha think?
The x-components are found by multiplying the magnitude of the vectors by the cosine of their respective angles, while the y components are found by multiplying the magnitude of the vectors by the sine of their respective angles.
Let's do the x-components for all the vectors first, so we get the x-component of the resultant vector:
[tex]F_{1x}=12 cos0[/tex] and
[tex]F_{1x}=12[/tex]
[tex]F_{2x}=9cos90[/tex] and
[tex]F_{2x}=0[/tex]
[tex]F_{3x}=15 cos126.87[/tex] and
[tex]F_{3x}=-9.0[/tex] (the angle of 126.87 is found by subtracting the 53.13 from 180, since angles are to be measured from the positive axis in a counterclockwise fashion).
That means that the x-component of the resultant vector, R, is 3.0
Now for the y-components:
[tex]F_{1y}=12sin0[/tex] and
[tex]F_{1y}=0[/tex]
[tex]F_{2y}=9sin90[/tex] and
[tex]F_{2y}=9[/tex]
[tex]F_{3y}=15sin126.87[/tex] and
[tex]F_{3y}=12[/tex]
That means that the y-component of the resultant vector, R, is 21.
Put them together in this way to find the resultant magnitude:
[tex]R_{mag}=\sqrt{(3.0)^2+(21)^2}[/tex] which gives us
[tex]R_{mag}=21[/tex] and now for the angle. Since both the x and y components of the resultant vector are positive, our angle will be where the x and y values are both positive in the x/y coordinate plane, which is Q1.
The angle, then:
[tex]tan^{-1}(\frac{21}{3.0})=82[/tex] degrees, and since we are QI, we do not add anything to this angle to maintain its accuracy.
To sum up: The resultant vector has a magnitude of 21 N at 82°
determine the work done when a forklift truck lifts a box of mass 350 kg a height of 2 m.
Answer:
work done = mgh
350×10×2
7000J
The work done by the forklift truck as it lifts the box to the given height is 6860J.
WorkWork is simply referred to as the displacement of an object when a push or pull force is applied to the object. It is the energy transferred from or to an object when force is applied to it along a displacement.
It is expressed as;
W = F × s
Where F is force and s is displacement
Given the data in the question;
Mass of box m = 350kgDisplacement s = 2mWork done W = ?We substitute our given values into the expression above.
W = F × s
But F = Weight = mass × acceleration due to gravity = mg
acceleration due to gravity ( g = 9.8m/s²)
Hence,
W = mg × s
W = 350kg × 9.8m/s² × 2m
W = 6860kgm²/s²
W = 6860J
Therefore, the work done by the forklift truck as it lifts the box to the given height is 6860J.
Learn more about Work: https://brainly.com/question/9942439
a car travel the first 20km with a speed of 40km/h and the next 40km with a speed of 80km/h . find the average speed
Answer:
average speed is 60km/h
Explanation:
you sum up the speed attained in each distance covered and divide it by 2 to get your answer
PLEASE HEEEEEEELP
Assume that the velocity of the soda bottle falling from a height of 0.8 m will be 4 m/s. Record this velocity for each mass in Table A, and use it in calculating the predicted kinetic energy of the soda bottle for the masses of 0.125 kg, 0.250 kg, 0.375 kg, and 0.500 kg using the equation: KE=1/2 mv^2 When solving for kinetic energy (KE), m is mass, and v is the speed (or velocity).
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
