Answer:
I think it's option D
Explanation:
I think it's option D but not so sure
When an object is in a gravitational field, it has energy in its __________ __________ energy store.
Answer:
Gravitational potential
Explanation:
Any object that is not on the surface of the Earth, but at a height instead has potential energy. Eventually, this can become kinetic energy once the object falls.
When an object is in a gravitational field, it has energy in its gravitational potential energy store.
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iron
22) How is it possible to fill medicine in a syringe?explain
I hope this helps you ^-^
The direction equivalent to {40° W of S} is:
A. 40 ° E of S
B. 40° W of N
C. 40° E of N
D. 50° S of W
E. 50° E of N
Answer:
c
Explanation:
Calculate the maximum absolute uncertainty for R if:
R = B - A
A = 32 +/- 2 seconds
B = 11 +/- 3 seconds
43 seconds
1 second
21 seconds
5 seconds
6 seconds
Answer:
ΔR = 5 s
Explanation:
The absolute uncertainty or error in an expression is
ΔR = | [tex]\frac{dR}{dB}[/tex] | ΔB + | [tex]\frac{dR}{dA}[/tex] | ΔA
the absolute value guarantees to take the unfavorable case, that is, the maximum error.
We look for the derivatives
[tex]\frac{dR}{dB}[/tex] = 1
[tex]\frac{dR}{dA}[/tex] = -1
we substitute
ΔR = 1 ΔB + 1 ΔA
of the data
ΔB = 3 s
ΔA = 2 s
ΔR = 3 + 2
ΔR = 5 s
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
A bus Starts from rest. If the acceleration of bus become 10 m/s2 after 15 sec Calculate the final Velocity of the bus
The mass of a brick is 2kg. Find the mass of water displaced by it when it is completely immersed in water. (Density of the bricks is 2.5 g/cm^3)
Answer:
2000g
Explanation:
volume=mass/density
=2000/2.5
=800cm³
mass=density×volume
=800×2.5
=2000g
I’m steel, the solvent is And the solute is. .
what will be the magnitude of work if a force of 25N pulls a stone through a distance of 5m in its direction?
Explanation:
125 is your answer........
. 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.
A steel ball is released just below the surface of thick oil in a cylinder.
During the first few centimetres of travel, what is the acceleration of the ball?
A constant and equal to 10 m / s2
B constant but less than 10 m / s2
C decreasing
D increasing
Answer:
Increasing
Explanation:
I Hope it Helps
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.
The boiling point of water is 1000 C at sea level. The boiling point of butane is -1.50C… If we leave liquid butane in a bowl on a table in a room where the temperature is 240C, butane will
A. evaporate.
B. condense.
C. freeze.
D. melt.
Answer: If we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.
Explanation:
A temperature at which the the liquid and gaseous phase of a substance of a substance are present in equilibrium with each other is called boiling point.
For example, the boiling point of butane is -1.5 degree Celsius.
This means that at a temperature above -1.5 degree Celsius, butane will exist is gaseous state. That is, at a temperature of 24 degree Celsius butane will evaporate.
Thus, we can conclude that if we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.
The gravitational force acting on various masses is measured on different planets. Measured values for the forces acting on the corresponding masses are shown in the data table. Analyze the data and develop a method for comparing the gravitational field strengths on the different planets. Use your method to compare the gravitational field strengths, and report your conclusions.
Question 4(Multiple Choice Worth 3 points)
(03.02 MC)
Which statement best reflects a change in weather?
Today is cloudy, but tomorrow will be clear and sunny.
The average rainfall has decreased over the past five years.
Ocean temperatures are projected to increase over time.
Glaciers are melting more rapidly now than in the past 100 years.
plzzz helpppp asap
Answer:
a is your answer.............
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°
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
when is the mass of an object if it exerts a force of 160 N and an acceleration of 8.15m/s^2
Answer:
f=ma.......m=f/a......m=20kg
Drag each label to the correct location on the image. Identify the particles and characteristics on this model of an atom.
Positively charged
Electron
Proton
Neutron
Negatively charged
Answer:
cant see picture
Explanation:
Answer:
please add picture so i can help you
Explanation:
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 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 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.
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
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.
¿Cuál de las siguientes no es un tipo de fuerza de roce
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]Example to measure the interval of time of a small stone dropped from 1m height.
Answer:
The time required is 0.45 s.
Explanation:
Height, h = 1 m
initial velocity, u = m/s
Let the time is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\1 = 0 +0.5 \times 9.8 \times t^2\\\\t = 0.45 s[/tex]
The summer camps had a field trip from the campus to Fragrance Hill. They traveled at an average speed of 65 km/h in the first 2 hours. After that, traveled at another average speed of 78 km/h. If the distance between the campus and Fragrance Hill is 364 km, what was the total time for the field trip?
Answer:
Explanation:
They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:
km/hr part 1 + km/hr part 2 = 364 km
Now we need to find each part on the left side of that equation. Part 1 first:
We traveled 65 km/hr for 2 hours, so that took us
[tex]65\frac{km}{hr}*2hr[/tex] and canceling out the hour label, we have that in part 1 we got
65(2) = 130 km. Good. Now onto the second part, where our unknown is.
We traveled 78 km/hr the second part for x hours, so that took us
[tex]78\frac{km}{hr}*xhr[/tex] and canceling out the hour label, we have that in part 2 we got
78x km. Now we can fill in the main equation (the one in bold print)
130 km + 78x km = 364 km and subtracting 130 km from both sides:
78x km = 234 km and dividing by 78 km:
x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.
An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image ,its nature and size?
Answer:
The position is 8.18cm from the mirror.
Nature is b=virtual
Size is 1.82cm
Explanation:
Note that for a convex mirror, the image distance and the focal length are negative;
Given
Object height H0 = 4cm
object distance u = 18cm
Radius of curvature R = 30cm
Since f = R/2
f = 30/2
f = -15cm
Recall that:
[tex]\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm[/tex]
Since the image distance is negative, this shows that the image is a virtual image.
To get the size:
[tex]\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm[/tex]
