Solution :
1. We know that : Buoyant force = weight of the liquid displace
= volume displaced x density of the fluid
Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]
Mass = weight / g
[tex]$=\frac{940}{9.8}$[/tex]
= 95.92 kg
And density = 1000 [tex]kg/m^3[/tex]
Therefore,
[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]
[tex]$=\frac{95.92}{1000}$[/tex]
= 0.0959 [tex]m^3[/tex]
We know density of air = 1.225 [tex]kg/m^3[/tex]
∴ Mass of air displaced = 0.0959 x 1.225
= 0.1175 kg
Weight of the air displaced = 1.1515 N
Therefore, the buoyant force = 1.1515 N
2). As the balloon is not accelerated, the net force acting on it is zero.
Thus the weight that acts downwards = buoyant force upwards
So, the weight of the air displaced = weight of the balloon
= 17000 N
Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air
[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]
[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]
= 1416.0766 [tex]m^3[/tex]
A baseball player (hitter) hits the ball with her bat. The ball flies north at an angle a above ground and with an initial speed v towards a fielder whose job is to catch the ball before it hits the ground. The distance between the fielder and the hitter is D. At the moment when the bat hits the ball, the fielder starts running to the south towards the place where he believes the ball will fall. Assume the fielder runs with a constant speed all the time the ball is in the air. Assume the hand of the fielder catches the ball just before it hits the ground and that the ball is hit by the bat at ground level.
How fast must the fielder run to be able to catch the ball?
Answer:
imma try and still get wrong
The fielder's required speed is (D + (v * t * cos(a))) / t.
To catch the ball, the fielder must run at a speed that allows them to cover the horizontal distance D between themselves and the hitter, as well as the distance the ball travels horizontally before hitting the ground. The time it takes for the ball to hit the ground can be determined based on its initial vertical velocity and the acceleration due to gravity.
Let's call the time the ball is in the air "t." The horizontal distance the ball travels is given by D + (v * t * cos(a)), where "v" is the initial speed of the ball and "a" is the angle of the ball's flight above ground. The fielder's running speed should equal this horizontal distance divided by "t" to ensure they reach the ball's landing point just in time to catch it. Thus, the fielder's required speed is (D + (v * t * cos(a))) / t.
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Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.
Required:
Rank from smallest to largest.
Answer:
Momentum of object A = Momentum of object C < momentum of B.
Explanation:
The momentum of an object is equal to the product of mass and velocity.
Object A has a mass m and a speed v. Its momentum is :
p = mv
Object B has a mass m/2 and a speed 4v. Its momentum is :
p = (m/2)×4v = 2mv
Object C has a mass 3m and a speed v/3. Its momentum is :
p = (3m)×(v/3) = mv
So,
Momentum of object A = Momentum of object C < momentum of B.
A covalent bond is formet by of electrons..?
Answer:
The covalent bond is formed by pairs of electrons that are shared between two atom
Explanation:
The covalent bond is formed by pairs of electrons that are shared between two atoms, in general the electrons must have opposite spins to have a lower energy state.
In this bond, the electrons are between the two atoms and are shared between them in such a way that there is a configuration of eight electrons in the orbit.
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
The answer is continuity ( D )
Explanation:
PLZ MARK AS BRAINLIEST
A spinning electron produces a(n)
a. element.
b. magnetic field.
c. proton.
d. piece of iron.
Answer:
A spinning electron produces a magnetic field.
Explains the fine structure of Hydrogen lines.
It is important to keep terms electrical potential and electrical potential energy straight. Let's say a charge q is placed at a point where the electrical potential is V. If I double the charge to 2q, the electrical potential of the point will _________ and the electrical potential energy of the charge will ___________ 1. stay the same; double 2. double; stay the same 3. stay the same; stay the same 4. double; double
Answer:
1. The electric potential would stay the same.
2. The electric potential energy would double
Explanation:
1. Since the electric potential at the point is V and is not dependent on the charge q, the electric potential at the point still remains the same even if the charge q is doubled to 2q.
So, the electric potential remains the same when the charge is doubled to 2q.
2. Since potential energy U = qV where q = charge and V = electric potential. So when the charge is q and electric potential, V, the electric potential energy at the point is U = qV.
When the charge is doubled to 2q and the electic potential remains the same, the electric potential energy is now U' = 2qV = 2U (since U = qV).
So, the electric potential energy at the point is doubled when the charge is doubled.
uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.00 cm. (a) What is the work done by the field on the electron? 1.753e-18 J (b) What is the change in potential energy associated with the electron? J
Answer:
a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J
Explanation:
a) work is defined by
W = F . x
the bold letters indicate vectors, in this case the force is electric
F = q E
we substitute
F = q E x
the charge of the electron is
q = - e
F = - e E x
let's calculate
W = - 1.6 10⁻¹⁹ 365 3 10⁻²
W = - 1.752 10⁻¹⁸ J
b) the change in potential energy is
U = q ΔV
the potential difference is
ΔV = - E. Δs
we substitute
U = - q E Δs
the charge of the electron is
q = - e
U = e E Δs
we calculate
U = 1.6 10⁻¹⁹ 365 3 10⁻²
U = + 1.752 10⁻¹⁸ J
What is the net force on the side of the container The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Answer:
26467.21 N
Explanation:
Initial height of water ( h1 ) = 5 m
diameter of container ( d1 )= 100 cm
pressure inside the container ( p1 )= 0.850 atm
Diameter of faucet ( d2 )= 1 inch
Calculate the value of the net force on the side of container
lets assume ; pressure outside the container ( p2 ) = 1 atm
Fnet = ( P1A1 + mg ) - ( P2A2 )
= [ ( 0.85 * 101325 ) ( π(1/2)^2 ) + mg ) - [ ( 101325 )( π )(0.0127)^2 ]
= [ 16902.2766 + pvg ] - [ 51.3161 ]
where ; pvg = pAhg = 1000 * π ( 1/2 )^2 * 5 * 9.8 = 9616.25
= [ 16902.2766 + 9616.25 ] - [ 51.3161 ] = 26467.21 N ( downwards )
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by a 12.0 - V battery, what total energy in Joules is delivered to the lightbulb filament during 2.00 s
Answer:
E = 20.03 J
Explanation:
Given that,
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,
Voltage, V = 12 V
We need to find the energy delivered to the lightbulb filament during 2.00 s.
The energy delivered is given by :
[tex]E=I^2Rt[/tex]. ....(1)
As,
[tex]I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A[/tex]
As per Ohm's law, V = IR
[tex]R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega[/tex]
Using formula (1).
[tex]E=0.835^2\times 14.37\times 2\\\\=20.03\ J[/tex]
So, the energy delivered to the lightbulb filament is 20.03 J.
A hungry monkey is sitting at the top of a tree 69 m above ground level. A person standing on the ground wants to feed the monkey. He uses a tee-shirt cannon to launch bananas at the monkey. If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the bananas, how should the person aim the banana cannon
Answer:
Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.
If the cannon throws the banana with the same force the monkey falls
(m.g=Fz <=> m.9,81N/kg=...N).
Then the throw will slow down because of the gravitational pull.
Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.
The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.
If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.
So to do so you need to throw the bananas with a speed of at least 9,81m.s
Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.
I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.
kind regards
A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? What is the amplitude?
The period and the amplitude of the weight suspended from spring are 0.33 seconds and 10 centimeters, respectively.
1) The period is given by:
[tex] T = \frac{1}{f} [/tex]
Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
[tex] T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s [/tex]
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
[tex] A = \frac{20 cm}{2} = 10 cm [/tex]
Therefore, the amplitude is 10 cm.
You can learn more about the period and amplitude here: https://brainly.com/question/15169209?referrer=searchResults
I hope it helps you!
HEELLPPPPPpppppppppppppppp
Explanation:
Given:
[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]
[tex]v_1[/tex] = 40 cm/s
[tex]v_2[/tex] = 90 cm/s
[tex]A_2[/tex] = ?
a) The continuity equation is given by
[tex]A_1v_1 = A_2v_2[/tex]
Solving for [tex]A_2[/tex],
[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]
[tex]= 2\:\text{cm}^2[/tex]
b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of
[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]
How many joules of energy are required to accelerate one kilogram of mass from rest to a velocity of 0.866c?
Answer:
the amount of energy needed is 1.8 x 10¹⁷ J.
Explanation:
Given;
mass of the object, m₀ = 1 kg
velocity of the object, v = 0.866 c
By physics convection, c is the speed of light = 3 x 10⁸ m/s
The energy needed is calculated as follows;
E = Mc²
As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;
[tex]M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\ M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg[/tex]
The energy required is calculated as;
E = 2 x (3 x 10⁸)²
E = 1.8 x 10¹⁷ J
Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.
How does radiation from the sun spread throughout Earth's atmosphere?
- through convection currents
- through conduction currents
- through solar panels
- through hot water systems
A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 16.4 m at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box
Over the first 16.4 m, the person performs
W = (60.0 N) (16.4 m) = 984 J
of work.
Over the remaining 6.88 m, they perform a varying amount of work according to
F(x) ≈ 60.0 N + (-8.72 N/m) x
where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :
W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J
(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)
So the total work performed by the person on the box is
984 J + 206.4 J = 1190.4 J ≈ 1190 J
write physical quantities and its unit
length= metre
mass= kg
time= second
temperature = kelvin
current= ampere
luminous intensity= candela
Amount of substance = mole
etc
I hope this will help you
stay safe
Una bala de 10 g se dispara contra un bloque de madera de 102 g inicialmente en reposo sobre una superficie horizontal. Después del impacto el bloque se desliza 8 m antes de detenerse. Si el coeficiente de fricción entre el bloque y la superficie es 0,5, ¿Cuál es la velocidad de la bala inmediatamente antes del impacto?
Answer:
una ess abola cola sola answer
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statement correctly describes the change in momentum of the two balls?
a. |ΔpBl<|ΔPA|
b. |ΔpBl=|ΔPA|
c. |ΔpB|>|ΔPA|
d. ΔpB > ΔPA
Answer:
Option A
Explanation:
From the question we are told that:
Mass [tex]m=0.20kg[/tex]
Velocity [tex]v=4m/s[/tex]
Generally the equation for momentum for Ball A is mathematically given by
Initial Momentum
[tex]M_{a1}=mV[/tex]
[tex]M_{a1}=0.2*4[/tex]
[tex]M_{a1}=0.8[/tex]
Final Momentum
[tex]M_{a2}=-0.8kgm/s[/tex]
Therefore
[tex]\triangle M_a=-1.6kgm/s[/tex]
Generally the equation for momentum for Ball B is mathematically given by
Initial Momentum
[tex]M_{b1}=mV[/tex]
[tex]M_{b1}=0.2*4[/tex]
[tex]M_{b1}=0.8[/tex]
Final Momentum
[tex]M_{b2}=-0 kgm/s[/tex]
Therefore
[tex]|\triangle M_a|>|\triangle Mb|[/tex]
Option A
A thin stream of water flows vertically downward. The stream bends toward a positively charged object when it is placed near it. The positively charged object is then removed. What will happen to the same stream of water when a negatively charged object is placed nearit
Answer:
the water jet and the negative object attract
Explanation:
Let's analyze the situation. Water is a good conductor of electricity, so when an object with a positive charge is brought closer, a charge of the opposite sign is created, which is why the two objects attract each other. The charge created comes from the ground
When claiming the object, the charge is reduced to zero or the charge goes to earth since water is a good conductor of electricity, therefore when approaching an object with a negative charge, more charges go to earth and the ring is left with a positive charge and attracts it to the object.
In short, the water jet and the negative object attract
Using the data provided below, calculate the corrected wavelength for a spectroscope reading of 6.32.
Spectroscope Readings Known Helium Wavelengths (nm)
3.65 388.9
4.75 486.6
4.90 501.6
5.65 587.6
6.45 667.8
7.00 706.5
Answer:
646.6 nm
Explanation:
Using the data given, we fit a linear model :
Spectroscope Readings Known Helium Wavelengths (nm)
3.65 388.9
4.75 486.6
4.90 501.6
5.65 587.6
6.45 667.8
7.00 706.5
Using technology like excel to fit the model, the regression model obtained is :
Y = 97.96582X + 27.48458
Where, y is the predicted or corrected wavelength value.
This mod could be used to calculate the corrected wavelength for a given Spectroscope value :
Given a Spectroscope value of 6.32 ; the corrected wavelength is obtained by replacing x in the equation by 6.32 and calculate y ;
Y = 97.96582X + 27.48458
x = 6.32
Y = 97.96582(6.32) + 27.48458
Y = 619.1439824 + 27.48458
Y = 646.6285624
Corrected wavelength value is : 646.6
A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.
a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______
Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]
(b) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]
El punto de ebullición de azufre es de 444.5°C.
c cual es la temperatura Correspondiente en la escala
Fahrenheit?
°F = 1.8°C + 32°
444.5°C = 832.1°F
For a solid uniformly charged sphere of radius R, calculate the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere. a) 9/8 b) infinity c) 2.0 d) 8.0 e) 8/9
Answer:
e
Explanation:
From the given information:
Suppose Q = total charge of the sphere.
here, the electric field outside the sphere at distance R/2 can be expressed as:
[tex]E_1 = \dfrac{1}{4 \pi \varepsilon _o}* \dfrac{Q}{(R + \dfrac{R}{2})^2}[/tex]
where:
[tex]k = \dfrac{1}{4 \pi \varepsilon _o}[/tex]
[tex]E_1 = \dfrac{kQ}{(\dfrac{3R}{2})^2}[/tex]
[tex]E_1 = \dfrac{4kQ}{9R^2}[/tex]
For the electric field inside the sphere, we have:
[tex]E_2 = \dfrac{kQr}{R^3}[/tex]
here:
r = distance of the point from the center = R/2
R = radius of the sphere
∴
[tex]E_2 = \dfrac{kQ * \dfrac{R}{2}}{R^3}[/tex]
[tex]E_2 = \dfrac{kQ }{2R^2}[/tex]
As such, the ratio of the electric field outside the sphere to the one inside is:
[tex]\dfrac{E_1}{E_2} = \dfrac{ \dfrac{4kQ}{9R^2}}{ \dfrac{kQ }{2R^2}}[/tex]
[tex]\dfrac{E_1}{E_2} = \dfrac{4kQ}{9R^2} \times \dfrac{ 2R^2 }{kQ}[/tex]
[tex]\mathbf{\dfrac{E_1}{E_2} = \dfrac{8}{9}}[/tex]
For a solid uniformly charged sphere of radius R, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is - e) 8/9
Electric field due to a sphere at a distance R/2 outside the sphere can be calculated by assuming the sphere as a point massfield [tex]E2= \frac{k(Q)}{(3R/2)^2}[/tex]
= [tex]\frac{4}{9} \frac{k(Q)}{R^2}[/tex]
The electric field at a distance R/2 inside the sphere E1= k(Q)(r)/R^3 where r is the distance from the center [tex]E1= \frac{k(Q)}{(3R/2)^3}[/tex]=
Thus, E2/E1
= (4/9)/(1/2)
= 8/9
Thus, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is - 8/9
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5a. i) Define pressure.
Answer:
Explanation:
How much force is acting on a certain area
The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?
a. WA = WB
b. WA = 2WB.
c. WA=4WB
d. WB= 2WA
Answer:
Option (c).
Explanation:
Let the mass of each cart is m and the force is F.
Time for cart A is 2t and for cart B is t.
Work done is given by the
W= force x displacement
As the distance is given by
S= u t +0.5 at^2
So, when the time is doubled the distance is four times.
So, WA = F x 4 S
WB = F x S
WA= 4 WB
Urgent please help !!!!!!!
Answer:
resultant vector =0
Explanation:
because it is connected head to tail in a closed figure
A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity
and deceleration are
and
Answer:
1. Initial velocity = 24 m/s
2. Deceleration = –4 m/s²
Explanation:
From the question given above, the following data were obtained:
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
1. Determination of the initial velocity.
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) =?
s = (u + v)t / 2
72 = (u + 0) × 6 / 2
72 = u × 3
Divide both side by 3
u = 72 / 3
u = 24 m/s
2. Determination of the deceleration.
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) = 24 m/s
Deceleration (a) =?
v = u + at
0 = 24 + (a × 6)
0 = 24 + 6a
Collect like terms
0 – 24 = 6a
–24 = 6a
Divide both side by 6
a = –24 / 6
a = –4 m/s²
What distance do I cover if I travel 10 m E, then 6 mW, then 12 m E?
A. 16 m
B. 28 m
C. 16 m E
D. 28 m E
Answer:
C. 16 m E
Explanation:
Applying,
The law of addition of vector: Vector in the same direction are added while vector in opposite direction are substracted
From the question above,
Step 1: Total distance covered towards east = 10+12 = 22 m E
Step2: Total distance covered towards west = 6 m W
Therefore, the resultant distance traveled = 22-6 = 16 m E
Hence the right option is C. 16 m E
An object carries a charge of -8.5 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the first to the second object so that both objects have the same charge?
Answer:
Approximately 2*10^13 electrons must be transferred
Explanation:
Below is the given information:
First object carries charge = -8.5 µC
Number of electrons in 1st = 8.5 x 10^-6/(1.6 x 10^-19) = 5.3125 x 10^13
Second object carries a charge = -2.0 µC
The number of electrons in 2nd = 2*10^-6/(1.6*10^-19) = 1.25 x 10^13
so, approximately 2 x 10^13 electrons must be transferred
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 3.22 x 104 V
Answer:
[tex]E=3.22*10^6 N/C[/tex]
Explanation:
From the question we are told that:
Separation Distance [tex]d=1.0cm =0.01m[/tex]
Potential difference [tex]V=3.22 * 10^4 V[/tex]
Generally the equation for Electric Field strength is mathematically given by
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{3.22*10^4}{0.01}[/tex]
[tex]E=3.22*10^6 N/C[/tex]
