Answer:
horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s
initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s
Using s = ut + at2/2 for change in vertical distance in time t, with acceleration a (-9.8m/s2) and initial velocity u (vv = 3m/s) we have
0 = 3*t - 9.8*t2/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).
The horizontal distance in time t is vh*t or 3√3*6/9.8 m
Explanation:
Ball 1 (1.5 kg) moves to the right at 2 m/s and ball 2
(2.5 kg) moves to the left at 1.5 m/s. The balls stick together after collision. What is the speed and direction of ball 2 after the collision?
Answer:
0.1875 m/s leftward
Explanation:
Taking rightwards as positive
We are given:
Ball 1:
Mass (m1) = 1.5 kg
velocity (u1) = 2 m/s
Ball 2:
Mass (m2) = 2.5 kg
velocity (u2) = -1.5 m/s [negative because it is in the opposite direction]
Speed and Direction of Ball 2:
We are told that the balls stick together after the collision
We can say that the balls have the same velocity since they are sticking together
So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s
According to the law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
replacing the variables
1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5) [v1 = v2 = V]
3 + (-3.75) = 4V
-0.75 = 4V
V = -0.75/4 [dividing both sides by 4]
V = -0.1875 m/s
Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction
A projector lens projects an image from a 6.35 cm wide LCD screen onto a
screen 3.25 m wide. If the focal length of the projector lens is 13.8 cm, the screen
must be how far from the projector
Answer:
For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.
20- A gram of distilled water at 4° C:
(a) will increase slightly in weight when heated to 6 C
(b) will decrease slightly in weight when heated to 6 C
(C) will increase slightly in volume when heated to 6 C
(d) will decrease slightly in volume when heated to 6 C
(e) will not change in either volume or weight
Answer:
D. will decrease slightly in volume when heated to 6° C
Explanation:
A gram of distilled water at 4° C will increase slightly in volume when heated to 6 C. Hence option C is correct.
What is Water ?Water has the chemical formula H2O, making it an inorganic substance. It is the primary chemical component of the Earth's hydrosphere and the fluids of all known living things (in which it serves as a solvent[1]). It is translucent, flavourless, odourless, and almost colourless. In spite of not supplying food, energy, or organic micronutrients, it is essential for all known forms of life. Its molecules are made up of two hydrogen atoms joined by covalent bonds and have the chemical formula H2O. The angle at which the hydrogen atoms are joined to the oxygen atom is 104.45°.[2] The liquid condition of H2O at normal pressure and temperature is known as "water" as well.
Water occurs because the environment on Earth is pretty near to the triple point of water.
To know more about Water :
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An ideal gas expands quasi-statically and isothermally from a state with pressurepand volumeVto a state with volume 4V. How much heat is added to the expanding gas?
Answer:
Q = PV(In 4)
Explanation:
We are told that the volume expands from V to a state with volume 4V.
Thus, initial volume is V and Final volume is 4V.
We want to find How much heat is added to the expanding gas.
For an isothermal process, the work done is calculated from;
W = nRT(In(V_f/V_i))
Where;
V_f is final volume
V_i is initial volume
Thus;
W = nRT(In(4V/V))
W = nRT(In 4)
Now, from ideal gas equation, we know that;
PV = nRT
Thus;
W = PV(In 4)
Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W
Where Q is quantity of heat
Thus;
Q = PV(In 4)
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
The mass of a paper-clip is 0.50 g and the density of its material is 8.0g/cm'. The total volume of
a number of clips is 20 cm.
How many paper-clips are there?
Answer:
320 paper clips
Explanation:
mass = volume × density = 20cm³ × 8g/cm³ = 160g
mass of 1 paper clip = 0.50g
mass of x paper clips = 160g
x = 160/0.50 = 320
A 1kg cannon ball.is fired horizontally with an initial velocity of 5m/s. If the cannon was atop a wall 20m above the ground, what is the total
change in KE?
Answer:
Ek = 196.2 [J]
Explanation:
The question concerns the KE kinetic energy.
That is, we must find the kinetic energy at the moment the cannon is fired and the kinetic energy of when the ball hits the ground after having fallen 20 meters.
At the moment when the ball is fired it is 20 meters above ground level. If the ground level is taken as the reference level of potential energy, where it is equal to zero, in this way when the ball is at the highest (20 meters) you have the maximum potential energy.
In this way, the energy in the initial state is equal to the sum of the kinetic energy plus the potential energy. As the energy is conserved this same energy will be present when the ball hits the ground, where the potential energy is zero and will have only kinetic energy.
[tex]E_{1}=E_{2}\\E_{k1}+E_{p1}=E_{k2}\\\frac{1}{2} *m*v^{2} +m*g*h=E_{k2}\\E_{k2}=0.5*1*(5)^{2} +1*9.81*20\\E_{k2}=208.7[J][/tex]
The kinetic energy in the initial state can be easily calculated by means of the following equation.
[tex]E_{k1}=\frac{1}{2} *m*v^{2}\\E_{k1}=0.5*1*(5)^{2}\\E_{k1}=12.5 [J][/tex]
Therefore the change in KE
[tex]E_{k} = 208.7 - 12.5\\E_{k} = 196.2 [J][/tex]
A tennis ball moves 18 meters northward, then 22 meters
southward, then 14 meters northward, and finally 28 meters
southward.
Answer:
The distance moved is 82 m.
The displacement is 18 m to the south.
Explanation:
The distance is a measure of the total length traveled along the path, while the displacement only takes into account the length between the starting position (departure) and final position (arrival). That is, distance refers to how much space an object travels during its movement, being the amount moved, while displacement refers to the distance and direction of the final position with respect to the initial position of an object.
So, the distance being the sum of the distances traveled, you get:
18 m + 22 m + 14 m + 28 m= 82 m
The distance moved is 82 m.
You know that the tennis ball moves 18 meters to the north, then 22 meters to the south, then 14 meters to the north, and finally 28 meters to the south. Then the tennis ball moves:
northward: 18 m + 14 m= 32 mto the south: 22 m + 28 m= 50 mCalculating the displacement as the difference between the final position and the initial position, you get:
displacement= 50 m - 32 m= 18 m
The displacement is 18 m to the south.
A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
Lisa throws a stone horizontally from the roof edge of a 50 meter high dormitory. It hits the ground at a point 60 m from the building. Find the time of flight.
Answer:
Explanation:
Time of flight is the time of takes to hit the ground
Given
Height H = 50m
Acceleration due to gravity g = 9.8m/s³
Using the equation of motion;
S = ut+1/2gt²
u = 0m/s
Substitute and get time t
50 = 0(t)+1/2(9.8)t²
50 = 4.9t²
t² = 50/4.9
t² = 10.204
t = √10.204
t = 3.19secs
Hence the time of flight is 3.19secs
Does anyone skateboard still?
Answer:
yes
Explanation:
people still skateboard that is an easy question
A freshly caught catfish is placed on a spring scale, and it oscillates up and down with a period of 0.19 s. If the spring constant of the scale is 2330 N/m, what is the mass of the catfish?
Answer:
The mass of the catfish is 2.13 kg
Explanation:
Period of oscillation, T = 0.19 s
spring constant, k = 2330 N/m
The period of oscillation of the spring is given by;
[tex]T = 2\pi \sqrt{\frac{m}{k} }\\\\\frac{T}{2\pi} = \sqrt{\frac{m}{k} }\\\\\frac{T^2}{4\pi^2} = \frac{m}{k}\\\\m = \frac{kT^2}{4\pi^2}[/tex]
where;
m is mass of the catfish
substitute the given values and solve for m;
[tex]m = \frac{kT^2}{4\pi^2} \\\\m = \frac{(2330)(0.19)^2}{4\pi^2} \\\\m = 2.13 \ kg[/tex]
Therefore, the mass of the catfish is 2.13 kg
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?) Express your answer to two significant figures and include the appropriate units.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
List Five examples from daily life in which you see periodic motion caused by a pendulum
(Marking Brainliest)
Answer:
by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.
Explanation:
Suppose a popular FM radio station broadcasts radio waves with a frequency of 96. MHz. Calculate the wavelength of these radio waves. Be sure your answer has the correct number of significant digits.
Answer:
3.125 meters.
Explanation:
(3.0*10^8)/(96*10^6)
= 3.125 meters.
Hope this helped!
A projectile is shot straight up from the earth's surface at a speed of 11,000 km/hr. How high does it go? ________km?
Taken from "Physics for Scientists and Engineers by Randall D. Knight 2nd Edition. Chapter 13 #34. There is an answer in the database already, but I do not understand it.
Answer:
476.35 km
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 11000 km/hr
Final velocity (v) = 0 km/hr (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) = ?
Next, we shall convert 9.8 m/s² to km/hr². This is illustrated below:
1 m/s² = 12960 km/hr²
Therefore,
9.8 m/s² = 9.8 m/s² × 12960 km/hr² / 1 m/s²
9.8 m/s² = 127008 km/hr²
Thus, 9.8 m/s² is equivalent to 127008 km/h²
Finally, we shall determine the maximum height reached by the projectile.
This is illustrated below:
Initial velocity (u) = 11000 km/hr
Final velocity (v) = 0 km/hr (at maximum height)
Acceleration due to gravity (g) = 127008 km/hr²
Maximum height (h) = ?
v² = u² – 2gh (since the projectile is going against gravity)
0² = 11000² – (2 × 127008 × h)
0 = 121×10⁶ – 254016h
Collect like terms
0 – 121×10⁶ = – 254016h
– 121×10⁶ = – 254016h
Divide both side by – 254016
h = – 121×10⁶ / – 254016
h = 476.35 km
Thus, the maximum height reached by the projectile is 476.35 km
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inches in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?
Answer:
Specific Gravity = 0.378
Explanation:
First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:
Pressure = Force/Area = Weight/Area
Weight = Pressure*Area
where,
Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²
Weight = (510 N/m²)(4.8 x 10⁻³ m²)
Weight = 2.48 N
Now, the weight is given as:
Weight = mg
2.48 N = m(9.8 m/s²)
m = (2.48 N)/(9.8 m/s²)
m = 0.25 kg
Now, we calculate volume of can:
Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)
Volume = 6.6 x 10⁻⁴ m³
Hence, the density of can will be:
Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³
Density of Can = 378.32 kg/m³
So, the specific gravity of Can will be:
Specific Gravity = Density of Can/Density of Water
Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)
Specific Gravity = 0.378
If the velocity of a car changes from 0 meters per second (m/s) to 100 m/s in 10 seconds, what is the acceleration over that 10 second period?
Answer:
10m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 100m/s
Time taken = 10s
Unknown:
Acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time.
A = [tex]\frac{v - u}{t}[/tex]
v = final velocity
u = initial velocity
t = time taken
So, insert the parameters and solve;
A = [tex]\frac{100 - 0}{10}[/tex] = 10m/s²
An explanation of the relationships among particular phenomena.
Answer:
Theory
Explanation:
Theory is a term that is used often in academic work or scientific research to explain certain things or conditions established on universal principles or laws.
It is used to describe the "why and how" or the reason behind the occurrence of a situation.
Hence, it is correct to conclude that THEORY is "an explanation of the relationships among particular phenomena."
Answer:
E) Theory
Explanation:
Edge 2020
Brainliest?
Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity
Answer: • They have same change in velocity but different changes in kinetic energy
•They started at the same height.
Explanation:
First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.
Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.
Therefore, based in the explanation, the correct options are:
• They have same change in velocity but different changes in kinetic energy
•They started at the same height.
If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.) Assume the specific heat of ice is 0.5
-30° C
40° C
-20° C
30° C
Answer:
-20°C
Explanation:
The specific heat capacity of ice using the cgs system is 0.5cal/g°C
The enthalpy change is calculated as follows
ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.
10cal = 2g×0.5cal/g°C×∅
∅=10cal/(2g×0.5cal/g°C)
∅=10°C
Final temperature= -30°C+ 10°C= -20°C
Answer:
-20 degrees Celsius
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which statement is not correct for lamps connected in parallel
Answer:
This question is not complete but the completed question is below
Which statement is not correct for lamps connected in parallel?
A They can be switched on and off separately.
B They will remain bright if another lamp is connected in parallel.
C They share the supply voltage equally between them.
D They still operate if one lamp is removed.
The correct option is A
Explanation:
Lamps connected in series have the same voltage running across each lamp in the connection and will thus have the same brightness if any lamp is added or removed. This property also means they can only be switched on and off by a single switch, hence option A is not correct about lamps connected in parallel.
Lamps connected in a parallel circuit will have the same voltage and different current.
A parallel circuit contains resistors arranged parallel to each other. some basic characteristics of parallel circuit include the following;
the voltage in all the resistors is the samethe current flowing in each resistor is differentV = I₁R₁ + I₂R₂ + I₃R₃ + ---
where;
V is the voltage in the circuitI is the different currentsR is the different resistorsThus, we can that lamps connected in a parallel circuit will have the same voltage and different current.
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An object in FREE-FALL on the MOON would experience which of the following
FORCES?
O a. Weight
O b. Normal
O c. Air Resistance
d. a and c
O e. None of these
Answer:
e. none of these
Explanation:
An object in FREE-FALL on the MOON would experience only acceleration
200. Un automóvil se desplaza
hacia la izquierda con
velocidad constante v,
en el momento en que se
deja caer un saco de lastre
desde un globo en reposo.
El vector que representa
la velocidad del saco vista
desde el automóvil en ese
instante en que se suelta es
Given F1: a force of magnitude 6 N at an angle of 30°
F2: a force of magnitude 8 N at an angle of 50°C
a. Find F1+ F2 analytically (using equations instead of graphing) and write it in the form Fr1i + Fr2 j
b. Find the magnitude FR and θ_resultant
Answer:
13.8 N
[tex]41.44^{\circ}[/tex]
Explanation:
[tex]F_1=6\ \text{N}[/tex]
[tex]F_2=8\ \text{N}[/tex]
[tex]F_1\cos\theta_1\hat{i}+F_1\sin\theta_1\hat{j}\\ =6\cos30^{\circ}+6\sin30^{\circ}\hat{j}\\ =5.2\hat{i}+3\hat{j}[/tex]
[tex]F_2\cos\theta_2\hat{i}+F_2\sin\theta_2\hat{j}\\ =8\cos50^{\circ}+8\sin50^{\circ}\hat{j}\\ =5.14\hat{i}+6.13\hat{j}[/tex]
[tex]F_R=F_1+F_2=10.34\hat{i}+9.13\hat{j}[/tex]
[tex]|F_R|=\sqrt{10.34^2+9.13^2}=13.8\ \text{N}[/tex]
The magnitude of the resultant is 13.8 N
Direction is given by
[tex]\tan^{-1}=\dfrac{y}{x}=\tan^{-1}\dfrac{9.13}{10.34}=41.44^{\circ}[/tex]
The angle of the resultant is [tex]41.44^{\circ}[/tex]
If you start at a speed of 4m/s and slow down to 2m/s in 4s what is your
acceleration?
Answer:
penis
Explanation:
A car is accelerated at a constant rate from 15 m/s to 25 m/s. It takes the car 6 s to reach its final speed. What is the car’s acceleration?
Answer:
1.67 m/s²Explanation:
The car’s acceleration can be found by using the formula
[tex]a = \frac{v - u}{t} \\ [/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question we have
[tex]a = \frac{25 - 15}{6} = \frac{10}{6} = \frac{5}{3} \\ = 1.666666...[/tex]
We have the final answer as
1.67 m/s²Hope this helps you
which of the following elements is the most reactive? Chlorine Bromine Fluorine Helium
Answer:
Fluorine is the most reactive
Explanation:
Among the halogens, fluorine, chlorine, bromine, and iodine, fluorine is the most reactive one. It forms compounds with all other elements except the noble gases helium (He), neon (Ne) and argon (Ar), whereas stable compounds with krypton (Kr) and xenon (Xe) are formed.
An airtight box, having a lid of area 80.0 cm^2, is partially evacuated. Atmospheric pressure is 1.01 Times 10^5 Pa. A force of 108 lb is required to pull the lid off the box. The pressure in the box was:_________.
Answer:
5×10^4Pa
Explanation:
Given force of 108 lb is required to pull the lid off the box,
To convert "Ib"to Newton ,we use conversation rate below
1 pounds = 4.4482216282509 newtons
Then 108 lb=x Newton
Cross multiply we have
X= 480.41Newton
The force that is needed to open the lid is F and pressure P.
We know that Pressure= Force/Area
Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then
80.0 cm^2 = 80×10^-4m^2
Substitute to the equation of the pressure we have
P= 480.41Newton/(80×10^4m^2)
P=6×10^4 Pa
The pressure in the box will be difference between the initial pressure and final pressure
=( 1.01 ×10^5 Pa)-(6×10^4 Pa)
= 50100Pa
= 5×10^4Pa
Therefore, The pressure in the box was
5×10^4Pa