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
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
49. A block is pushed across a horizontal surface with a
coefficient of kinetic friction of 0.15 by applying a
150 N horizontal force.
(a) The block accelerates at the rate of 2.53 m/s2
Find the mass of the block.
(b) The block slides across a new surface while
experiencing the same applied force as before.
The block now moves with a constant speed.
What is the coefficient of kinetic friction between
the block and the new surface?
Answer:
(a) 37.5 kg
(b) 4
Explanation:
Force, F = 150 N
kinetic friction coefficient = 0.15
(a) acceleration, a = 2.53 m/s^2
According to the newton's second law
Net force = mass x acceleration
F - friction force = m a
150 - 0.15 x m g = m a
150 = m (2.53 + 0.15 x 9.8)
m = 37.5 kg
(b) As the block moves with the constant speed so the applied force becomes the friction force.
[tex]F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4[/tex]
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
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.
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
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.
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
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
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.
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.
Compared with dim light, what do light waves that look bright tend to have the subject just says science but the picker doesn't have that
Answer:
The brightness of a light depends on the amplitude of the light wave, which is the extent the waves moves from their equilibrium position. The brightness is also related to the amount of light that is emitted or reflected by an object
Therefore, compared to dim light, light that look bright have a higher amplitude and emit or reflect more light energy (photons)
Explanation:
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.
22) How is it possible to fill medicine in a syringe?explain
I hope this helps you ^-^
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.
An egg is dropped onto a wood floor and breaks. When a different egg (same mass) is dropped the same distance onto a foam pad on the wood floor, it does not break. Why (in terms of physics) does this happen?
A. The pad extends the time so the impulse changes.
B. An egg requires a harder surface so it won’t break.
C. The pad pulls the egg to the floor.
D. Wood is harder than the pad.
Answer:
A.
Explanation:
The pad extends the time so the impulse changes.
Answer:
A. The pad extends the time so the impulse changes
Explanation:
in term of physics it has same mass:)
Which of the following travels with a wave?
O A. Energy only
O B. Matter only
O C. Neither matter nor energy
O D. Both matter and energy
Answer:
A. Energy only is the answer to your question
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.............
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
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.
2 What are(i) free fall , (ii) acceleration due to gravity, (iii) escape velocity , (iv) centripetal force?
Answer:
Explanation:
1. Free fall implies an object falling under the gravitational influence only. During the flight, no other force acts on it except the gravitational pull.
2. Acceleration due to gravity is the earth's natural force of pull on all objects on its surface, close to its surface, or in the region where the force can be felt. This force pulls object to the surface of the earth.
3. Escape velocity is the required minimum velocity for an object to leave the gravitational influence of the earth. It has a constant value which can be determined by;
Escape velocity = [tex]\sqrt{2gR}[/tex]
where g is the gravitation force of the earth and R is the radius of the earth.
4. Centripetal force is the force of pull that is required to keep a rotation object in its curved path.
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
a uniform meter ruler is balanced at its midpoint
Answer:
a) i) x = 0.25 m, ii) x = 0.10 m, iii) x = 0.050 m
b) i) x = 0.40 m
Explanation:
a) For this exercise we use the rotational equilibrium equation, where we assume that the anticlockwise rotations are positive.
1) L = 2W
we set our reference system in the center of the bar where the fulcrum is
∑τ = 0
W 0.50 - L x = 0
x = 0.50 W / L
we substitute the value
x = 0.50 W / 2W
x = 0.25 m
ii) L = 5W
we calculate
x = 0.50 W / 5W
x = 0.10 m
iii) L = 10 W
x = 0.50 W / 10W
x = 0.050 m
b) a new weight is placed at x₂ = 30 cm on the left side
W 0.50 + W 0.30 - L x = 0
x = (0.50 + 0.30) W / L
x = 0.80 W / L
we calculate
i) L = 2W
x = 0.80 w / 2w
x = 0.40 m
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]2(a)
A car accelerates from 22ms^-1 to 43ms*-1 in 18.6 seconds.
(a) Calculate the acceleration of the car.
2(b)
Find the distance covered by the car in 10 seconds.
2(c) Find the velocity after 5 seconds.
Answer:
2a.
a=1.13ms^-2
2b.
S=277m
2c.
V=27.7ms-¹
Explanation:
Initial Velocity (U)=22m/s¹
Final Velocity (V)=43m/s²
Time(t) =18.6s
a. a=V-U/t
a=43-22/18.6
a=1.129
a=1.13m/s²
2b.
S=ut+1/2 at²
s=22(10)+1/2×1.13(10)²
s= 220+0.57(10)²
s= 220+0.57(100)
s= 220+57
s=277m
2c.
V=U+AT
V=22+1.13(5)
V=22+5.65
V=22+5.7
V=27.7m/s¹
ANSWER ASAP What happens to a circuit’s resistance (r) , voltage (v) , and current (l) when you decrease the length of the wire in the circuit ?
A . R decreases
V constant
l increases
B . R constant
V increases
l increases
C . R increases
V constant
l decreases
D . R increases
V decreases
I decreases
Answer:A
Explanation:i just took the test
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
A cricket player lowers his hands while catching the ball. wny?
. 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 bus Starts from rest. If the acceleration of bus become 10 m/s2 after 15 sec Calculate the final Velocity of the bus
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°
