Answer:
The
Explanation:
The horizontal distance covered is known as the range expressed as;
R = U√2H/g
U is the speed = 200m/s
H is the max height = 800m
g is the gravity = 9.8m/s²
R = 200√2(800)/9.8
R = 200√1600/9.8
R = 200√163.27
R = 200(12.78)
R = 2555.54
Hence the package should be released at a distance of 2555.54m
A bullet of mass 4.2 g strikes a ballistic pendulum of mass 1.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.
Answer:
449.020m/s
Explanation:
We have been provided with the following information:
Mass of bullet = Mb = 4.2g
Mass of pendulum = 1.0 = Md
Vertical distance = h = 18cm
Gravitational force = g = 9.8
We have kinetic energy converted to potential energy for the entire system
1/2(Mb+Md)V² = (Mb + Md)gh
We have V as the speed of system during collision
Mbv = (Mb+Md)V
We divide through by Mb
v = (Mb+Md/Mb)√2gh
4.2x10^-3 = 0.0042
18cm = 0.18m
(0.0042+1.0/0.0042)√2x9.8x0.18
= 1.0042/0.0042√3.528
= 239.095x1.878
= 449.020m/s
This is the answer to the question
Thank you!
What is the volume of a box if he has Length=7 cm Width=5cm , Height=10cm ?
Answer:
Volume of Cuboid = Height*Width*Length
Explanation:
Volume of Cuboid = 10*5*7
= 350 cu² cm
Answer:
Diagram:-[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]
Required Answer:-It is a cuboid
where
length =l=7cmwidth=b=5cmheight =h=10cmAs we know that in a cuboid
[tex]{\boxed{\sf Volume=lbh}}[/tex]
Substitute the values[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]
A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy
Which statement best describes a characteristic of gases?
Gases can be compressed, or squeezed together.
The particles of gases are packed close together.
The particles of gases spread our vertically instead of horizontally.
Gases have a definite shape and volume.
Answer:
A
Explanation:
I'm pretty sure it's A. That's the one that makes the most sense and checks out
Answer:
Just here to confirm that it is A
Explanation:
what is gathering and analyzing information about an object without physical contact with the object
Answer:
Remote Sensing
Explanation:
The speed of a space shuttle is 10 / express this in /�
Answer:
268.22m/s
Explanation:
Given;
10mile/min to m/s
We need to convert between the two units;
1 mile = 1609.34m
60s = 1min
Now;
10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]
= 268.22m/s
The sound intensity at 4 m from a source is 100 W/me. What is the intensity of the sound at 12 m away from the source ?
Answer:
Intensity at 12 meters will be 11.11 W/m^2
Explanation:
Recall that the intensity of sound is inversely proportional to the square of the distance from the source. Therefore, if at 4 m the intensity is 100 W/m^2
we have: 100 W/m^2 = k/16 and therefore, k = 1600 W
Then the intensity (I) at 12 m will be:
I = k/12^2 = 1600/144 W/m^2 = 11.11 W/m^2
A 8.45μC particle with a mass of 6.15 x 10^-5 kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m. How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
This question is incomplete, the complete question is;
A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.
How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
Answer:
it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
Explanation:
Given that;
mass m = 6.15 x 10⁻⁵ kg
q = 8.45μC = 8.45 × 10⁻⁶ C
B = 0.493
we know that
Time period T = 2πr / V
where r = mv/qB
so T = 2πm/qB
we substitute
T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)
T = 0.0003862 / 0.000004165
T = 92.7 sec
Therefore it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why
Answer:
hsvshxansjusjsnwjwisks
Explanation:
When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..
Sona wants to install room heater in her living room. She had only two options, either to install heater at the top of the window or near the ground level. . Which method of installing room heater would be the effective way . Why or Why not ?
A uniform electric field has a magnitude of 10 N/C and is directed upward. A charge brought into the field experiences a force of 50 N downward. The charge must be:_______.
Answer:
q = 5 C
Explanation:
The electric field is defined as the force experienced by a unit charge when it is brought into the field. Hence, the formula used to find the electrical field is given as follows:
E = F/q
where,
E = Electric Field Magnitude = 10 N/C
F = Force Experienced by the test charge = 50 N
q = Magnitude of the Charge = ?
Therefore,
10 N/C = 50 N/q
q = 50 N/(10 N/C)
Therefore,
q = 5 C
An extraterrestrial creature is standing in front of plane mirror. The height of this creature is H and we know that this creature has eyes positioned h below the top of its head. This creature sees its reflection which fit exactly the mirror, it means, this creature can just see the top of head and the bottom of its feet (or whatever it uses for motion). We can conclude that the top of a mirror is exactly:________
a. H/2 above the ground
b. H above the ground
c. (H-h/2) above the ground
d. (H-h) above the ground
e. We can not guess anything without information about the nature of this creature.
Answer:
c. (H-h/2) above the ground
Explanation:
The mirror must be at least half as tall as the alien, and its base must be located at half of the distance between the alien's eyes and the ground (assuming that the alien doesn't float or levitate).
This question is about the Law of Reflection which states that the angle of reflection = angle of incidence.
I attached an image that can help you understand the concept, although the alien is not included.
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.Express your answer using two significant figures.m venus atmosphere = ____ kg
Answer:
m = 4.9 10⁸ kg
Explanation:
The expression for the density is
ρ = m / V
m = ρ V
the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant
V = V_atmosphere - V_planet
V = 4/3 π R_atmosphere³ - 4/3 π R_venus³
V = 4/3 π (R_atmosphere³ - R_venus³
)
the radius of the planet is R_venus = 6.06 10⁶ m.
The radius of the outermost layer of the atmosphere
R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶
R_atmosphere = 6.11 10⁶ m
let's find the volume
V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]
V = 23,265 10⁶ m³
let's calculate the mass
m = 21 23,265 10⁶
m = 4.89 10⁸ kg
with two significant figurars is
m = 4.9 10⁸ kg
What physical property does the symbol I_enclosed in problem 5 represent? a. The current along the path in the same direction as the magnetic field b. The current in the path in the opposite direction from the magnetic field c. The total current passing through the loop in either direction d. The net current through the loop
Answer:
C
Explanation:
Current passing through the loop in either direction
Objects falling through air are slowed by the force of air resistance. Which objects were slowed the most by air resistance?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, it should be noted that when objects of different sizes fall in absence of air resistance, the objects will get to the ground at the same time. But with the presence of air resistance, the heaviest object gets to the ground first; meaning it has the least air resistance while the lightest object will arrive at the ground last because it has the greatest air resistance and is slowed down the most by the air resistance.
Thus, the lightest object in the completed question is the answer.
A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value?(c) What is that maximum value?
Answer:
A) r = 0.03 m
B) r = 0.0533 m
C) B_max = 0.00003 T
Explanation:
Formula for magnetic field inside the capacitor when it is parallel to the length element is;
B_in = (μ_o•I•r/(2πR²)
Formula for maximum magnetic field is;
B_max = (μ_o•I/(2πR)
Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
A) Magnetic field inside the capacitor is gotten from our first equation above;
B_in = (μ_o•I•r/R²)
Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.
Thus;
B_in = 0.75B_max
(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))
μ_o•I and 2πR will cancel out to give;
r/R = 0.75
r = 0.75R
We are given R = 40 mm = 0.04 m
r = 0.75 × 0.04
r = 0.03 m
B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:
B_out = 0.75B_max
(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))
μ_o•I and 2π will cancel out to give;
1/r = 0.75/R
r = R/0.75
r = 0.04/0.75
r = 0.0533 m
C) B_max = μ_o•I/(2πR)
μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A
Thus;
B_max = (4π × 10^(-7) × 6)/(2π × 0.04)
B_max = 0.00003 T
If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)
Question 10 options:
There cannot be any forces applied to the ball.
There must be exactly one force applied to the ball.
The net force applied to the ball is zero.
The net force applied to the ball is directed to the right.
Answer:
C. The net force applied to the ball is zero.
Explanation:
From Newton's second law of motion;
F = ma
Where F is the force on an object, m is its mass and a is its acceleration.
Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.
Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.
So that;
F = m x 0
= 0
No force is applied on the object.
Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.
We will now determine the indexes of refraction for two Mystery materials, A and B. These materials can be selected from the list of materials on the right. Be sure to set your laser pointer to a frequency of 589 nm. Questions:A. Devise an experiment for determining the indices of refraction for these. Explain your methodology. B. What are the indices of refraction for the two mystery materials, A and B?
Answer:
A) refraction experiment n = n₁ sin θ₁ / sin θ₂
B) n_A = 1.19 , n_B = 1.53
Explanation:
A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index
n₁ sin θ₁ = n₂ sinθ₂
n₂ = n₁ sin θ₁₁ /sin θ₂
If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is
n = n₁ sin θ₁ / sin θ₂
B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be
material A
n_A = sin 50 / sin 40
n_A = 1.19
material B
n_B = sin 50 / sin30
n_B = 1.53
In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special device was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.
I really don't know how to do any of this please help me :(
Answer:
V₀ = 45.81 m/s
H = 70.45 m
T = 5.36 s
Explanation:
The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 201.24 m
V₀ = Initial Speed = ?
θ = Launch Angle = 35°
g = 9.8 m/s²
Therefore,
201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²
V₀ = √[(201.24 m)/(0.095 m/s²)
V₀ = 45.81 m/s
Now, for maximum height:
H = V₀² Sin² θ/g
H = (45.81 m/s)² Sin² 35°/9.8 m/s²
H = 70.45 m
For the total time of flight:
T = 2 V₀ Sin θ/g
T = 2(45.81 m/s) Sin 35°/9.8 m/s²
T = 5.36 s
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 71.0 kg, and the height of the water slide is 12.7 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide
Answer:
[tex]10.27m/s[/tex]
Explanation:
Given data
work W= -5.10 10^3 J
mass m= 71kg
final height of slide h2= 12.7m
initial height of slide h1=0m
initial velocity v1= 0m/s
final velocity v2=?
Step two:
required
Final velocity
The work-energy theorem is expressed as'
[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]
make V2 subject of formula we have final speed
[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]
substitute our given data we have
[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]
The student going at 10.27m/s
Any five physics problems
Explanation:
There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.
On a surface of a planet of radius R and mass M the acceleration due to gravity is 7m/s?. Consider another planet of radius 2R and mass 0.4M. What would the acceleration due to gravity be on this new planet? Show your calculations.
Answer:
0.7 m/[tex]s^{2}[/tex]
Explanation:
From Newton's law of universal gravitation,
F = [tex]\frac{GMm}{r^{2} }[/tex]
and from Newton's second law of motion,
F = mg
So that;
mg = [tex]\frac{GMm}{r^{2} }[/tex]
⇒ g = [tex]\frac{GM}{r^{2} }[/tex]
For the first planet,
7 = [tex]\frac{GM}{R^{2} }[/tex]
⇒ G = [tex]\frac{7R^{2} }{M}[/tex] .............. 1
For the second planet,
g = [tex]\frac{0.4GM}{(2R)^{2} }[/tex]
= [tex]\frac{0.4GM}{4R^{2} }[/tex]
⇒ G = [tex]\frac{4gR^{2} }{0.4M}[/tex] ............. 2
Equating 1 and 2, we have;
[tex]\frac{7R^{2} }{M}[/tex] = [tex]\frac{4gR^{2} }{0.4M}[/tex]
g = [tex]\frac{7R^{2} *0.4M}{4R^{2}M }[/tex]
= [tex]\frac{7*0.4}{4}[/tex]
= [tex]\frac{2.8}{4}[/tex]
g = 0.7
Therefore, the acceleration due to gravity on the new planet is 0.7 m/[tex]s^{2}[/tex].
A 500-N box is at rest on the floor. Dennis Elbo makes several
attempts to move the box, pushing against the box with varying
amounts of horizontal force. Yet the box never does move. In this
situation, the amount of static friction force experienced by the box
Select all that apply.
-
0 is 500 N
O is equal to the force with which Dennis exerts on the
box
has an upper limit and Dennis O has not yet exceeded the upper limit
Ois always the coefficient of friction multiplied by the normal force value
Answer:
Select the second and the third options you listed.
Explanation:
Select the answer options:
"is equal to the force with which Dennis exerts on the box."
and
"has an upper limit and Dennis has not yet exceeded the upper limit."
In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.
explain the relationship among visible light, the electromagnetic spectrum, and sight.
Explanation:
The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.
Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.
Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)
Answer:
the object's mass is 50 kg
Explanation:
We use Newton's second law to solve for the mass:
F = m * a , then m = F / a
In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:
m = w / a = 650 N / 13 m/s^2 = 50 kg
Then, the object's mass is 50 kg.
The components of lifetime fitness include all of the following components except
Answer:it’s A
Explanation:
because i took the quiz
Answer:
D is the correct answer, not A
Explanation:
What percentage of an iron anchor’s weight will be supported by buoyant force when submerged in salt water?
Answer:
0.87
Explanation:
To solve this, we use the principle of Archimedes. Archimedes Principal of flotation states that "the buoyant force of an object is equal to the total weight of the fluid it displaces."
In the attachment, I stated the mathemacal formula, of which
F(B) = The buoyant force
w(fl) = The weight of the salt water displaced
p(iron) = density of iron
p(salt) = density of the salt water = 1025 kg/m³
F' = weight of the iron in air
F = weight of the iron in salt water
p(man) = density of man = 7680 kg/m³
The rest are the easy calculations done by substituting the values
F = 5 Newtons
W = 75 Joules
d = ?
ANSWER
A current of 3.75 A in a long, straight wire produces a magnetic field of 2.61 μT at a certain distance from the wire. Find this distance.
Given :
Current, I = 3.75 A .
Magnetic Field, [tex]B = 2.61\times 10^{-4}\ T[/tex]
To Find :
The distance from the wire.
Solution :
We know,
[tex]B = K\dfrac{2i}{d}\\\\d = 10^{-7}\times \dfrac{2\times 3.75}{2.61\times 10^{-4}}\\\\d = 0.00287\ m \\\\d = 2.87\times 10^{-3}\ m[/tex]
Hence, this is the required solution.