Closer to the sun . . . orbital speed is faster.
Farther from the sun . . . orbital speed is slower.
Flag answer: Answer 13 Answer 13
A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the spring decompresses and returns to its equilibrium length, causing the ball to speed up, before the ball enters the horizontal barrel of the cannon. The horizontal barrel is 15.0 cm long and it exerts a constant friction force of 0.032 N on the ball. A. With what speed does the projectile leave the barrel of the cannon
Answer:
1.40 m/s
Explanation:
The potential energy of a compressed spring can be expressed as:
[tex]E_{ps}=\dfrac{1}{2}kx^2[/tex]
From above;
k = spring constant
x = distance of the spring (compressed)
From the barrel, the kinetic energy (i.e. the final K.E) of the ball is calculated using the relation:
[tex]E_{kf}= \dfrac{1}{2}mv^2[/tex]
where;
m = the ball mass
v = ball's speed
Equating both equations above, we have:
[tex]E_{ps}- F_fd=E_{kf[/tex]
This can be re-written as:
[tex]\dfrac{1}{2}kx^2 - F_fd=\dfrac{1}{2}mv^2}[/tex]
[tex]v^2 = (\dfrac{k}{m})x^2-\dfrac{2F_fd}{m}[/tex]
[tex]v =\sqrt{ (\dfrac{k}{m})x^2-\dfrac{2F_fd}{m}}[/tex]
replacing the values from the given information:
[tex]v =\sqrt{ (\dfrac{8.00\ N/m}{5.30\times10^{-3} \ kg})(5.00 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm})^2-(\dfrac{2(0.032 \ N)(0.150 \ m)}{5.30\times \dfrac{10^{-3} \ kg}{1 \ g}})}[/tex]
[tex]v = \sqrt{1.962264151}[/tex]
v ≅ 1.40 m/s
The speed at which the projectile leaves the barrel of the cannon will be given as [tex]v=1.40\ \frac{m}{s}[/tex]
What is speed?Speed is defined as the movement of any object with respect to time. It is the ratio of distance and time.
Now it is given in the question:
Mass of ball m = 5.30 g
The deflection of spring = 5 cm
The force constant of spring [tex]k= 8 \ \frac{N}{m^2}[/tex]
The length of the barrel = is 15 cm
The frictional force of the barrel = 0.032 N
Now from the conservation of energy, we can write as
[tex]E_{spring}-E_{friction}=E_{ball}[/tex]
[tex]\dfrac{1}{2} kx^2-F_fd=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{\dfrac{k}{m}(x^2) -\dfrac{2F_fd}{m} }[/tex]
Now putting the values in the above formula:
[tex]v=\sqrt{\dfrac{8}{5.30\times 10^{-3}}(15\times10^{-2}) -\dfrac{2\times(0.0032)\times (0.015)}{5.30\times 10^{-3}} }[/tex]
[tex]v=1.40\ \frac{m}{s}[/tex]
Thus the speed at which the projectile leaves the barrel of the cannon will be given as [tex]v=1.40\ \frac{m}{s}[/tex]
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YO giving 49 dollars if you answer this question .
. ( )
Trace the flow of energy through the food pyramid, label the missing units of energy found at each level by calculating the change in energy
Answer:
Energy Flow
The chemical energy of food is the main source of energy required by all living organisms. This energy is transmitted to different trophic levels along the food chain. This energy flow is based on two different laws of thermodynamics:
First law of thermodynamics, that states that energy can neither be created nor destroyed, it can only change from one form to another.
Second law of thermodynamics, that states that as energy is transferred more and more of it is wasted.
Energy Flow in Ecosystem
The energy flow in the ecosystem is one of the major factors that support the survival of such a great number of organisms. For almost all organisms on earth, the primary source of energy is solar energy. It is amusing to find that we receive less than 50 per cent of the sun’s effective radiation on earth. When we say effective radiation, we mean the radiation, which can be used by plants to carry out photosynthesis.
Explanation:
Energy Flow in Ecosystem
The energy flow in the ecosystem is one of the major factors that support the survival of such a great number of organisms. For almost all organisms on earth, the primary source of energy is solar energy. It is amusing to find that we receive less than 50 per cent of the sun’s effective radiation on earth. When we say effective radiation, we mean the radiation, which can be used by plants to carry out photosynthesis.
Most of the sun’s radiation that falls on the earth is usually reflected back into space by the earth’s atmosphere. This effective radiation is termed as the Photosynthetically Active Radiation (PAR).
Overall, we receive about 40 to 50 percent of the energy having Photosynthetically Active Radiation and only around 2-10 percent of it is used by plants for the process of photosynthesis. Thus, this percent of PAR supports the entire world as plants are the producers in the ecosystem and all the other organisms are either directly or indirectly dependent on them for their survival.
The energy flow takes place via the food chain and food web. During the process of energy flow in the ecosystem, plants being the producers absorb sunlight with the help of the chloroplasts and a part of it is transformed into chemical energy in the process of photosynthesis.
This energy is stored in various organic products in the plants and passed on to the primary consumers in the food chain when the herbivores consume (primary consumers) the plants as food. Then conversion of chemical energy stored in plant products into kinetic energy occurs, degradation of energy will occur through its conversion into heat.
Then followed by the secondary consumers. When these herbivores are ingested by carnivores of the first order (secondary consumers) further degradation will occur. Finally, when tertiary consumers consume the carnivores, energy will again be degraded. Thus, the energy flow is unidirectional in nature.
Moreover, in a food chain, the energy flow follows the 10 percent law. According to this law, only 10 percent of energy is transferred from one trophic level to the other; rest is lost into the atmosphere. This is clearly explained in the following figure and is represented as an energy pyramid.
When converting chemical potential energy to kinetic energy some of the energy is lost as heat.
Answer:
When converting chemical potential energy to kinetic energy some of the energy is lost as heat.
Choose the smallest item from the list below.
1 glass of water
1 droplet of water
1 atom of oxygen
1 molecule of water
Answer
One molecule of water
Q2 A source of frequency 500 Hz emits waves of
wavelength 0.2m. How long does it take the waves to
travel 400m?
Answer:
4 secs
Explanation:
The first step is to calculate the velocity
V= frequency × wavelength
= 500× 0.2
= 100
Therefore the time can be calculated as follows
= distance/velocity
= 400/100
= 4 secs
The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doubled, the new force would have what magnitudes?
Answer:
F' = F
Explanation:
The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:
[tex]F = \frac{Gm_1m_2}{r^2}[/tex]
where,
F = Force of attraction
G = Universal gravitational costant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects
Now, if the masses and the distance between them is doubled:
[tex]F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}[/tex]
F' = F
A volleyball is served at a speed of 8 / at an angle 35° above the horizontal. What is the speed of the ball when received by the opponent at the same height?
3. How can a generator that otherwise produces AC
current be modified to produce DC current?
A proton accelerates from rest in a uniform electric field of 595 N/C. At one later moment, its speed is 1.15 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?
A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit is 2.0 hours. The radius of the planet is 4.29 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?
Answer:
The weight of the planet is 29083.5 N .
Explanation:
mass of satellite, m = 6463 kg
height of orbit, h = 4.82 x 10^5 m
period, T = 2 h
radius of planet, R = 4.29 x 10^6 m
Let the acceleration due to gravity at the planet is g.
[tex]T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2[/tex]
The weight of the satellite at the surface of the planet is
W = m g = 6463 x 4.5 = 29083.5 N
What does this circle graph tell you about water on Earth? (2 points)
a pie graph with a big blue section covering seventy one percent and small gray section covering twenty nine percent with a key indicating that blue is water and gray is land
Fresh water covers 71 percent of Earth's surface.
Oceans covers 71 percent of Earth's surface.
Salt water covers 71 percent of Earth's surface.
Water covers 71 percent of Earth's surface.
Answer:
ocean covers 71 percent of the earth
Answer:
the ocean covers 71 percent of Earth's surface.
Explanation:
A truck has a mass of 1.5 x 104 kg. If the truck can reach a maximum acceleration of 1.5 m/s2, what is the net force the truck exerts?
2.25 x 105 N
2.25 x 104 N
2.3 x 104 N
2.3 x 105 N
Answer:
2.25 × 10⁴ NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 1.5 × 10⁴ × 1.5
We have the final answer as
2.25 × 10⁴ NHope this helps you
SOMEONE HELP ME PLSS
Answer:
Explanation:
1 minute =60 second
power=work done/time taken
=2500/60
=41.66 watt
work=force *displacement
=500 N * 100 m
=50000 joule
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A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collision, the car travels 2.0m before stopping. Determine the resultant of the forces acting on the car.
Answer:
Explanation:
The vehicle is experiencing a large force created by the concrete wall.
Equation
vf^2 = vi^2 + 2*a * d
Givens
vf = 0 The car eventually does stop.
vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]
vi = 20 meters / second
a = ?
m = 850 kg
Solution
vf^2 = vi^2 + 2a*d
0 = 20 m/s + 2* 2 *a
-20 m/s = 4a
-20/4 = a
a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.
Force = m * a
F = - 850 * (-5)
F = - 4250 N
The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.
A steel playground slide is 5.25 m long and is raised 2.75 m on one end. A 45.0 kg child slides down from the top starting at rest. The final speed of the child at the bottom is 6.81 m/s. Find the average force of friction between the child and the slide.
Answer:
[tex]F=32.24N[/tex]
Explanation:
From the question we are told that:
Height [tex]h= 2.75 m[/tex]
Length[tex]l = 5.25 m[/tex]
Mass [tex]m=45kg[/tex]
Final speed [tex]v_f=6.81[/tex]
Generally the equation for Potential Energy P.E is mathematically given by
[tex]P.E=mgh[/tex]
Therefore
Initial potential energy
[tex]P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=0.5mv^2[/tex]
Therefore
Final kinetic energy
[tex]K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J[/tex]
Generally the equation for Work_done is mathematically given by
[tex]W=P.E_1-K.E_2\\\\W=169.3[/tex]
Therefore
[tex]F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}[/tex]
[tex]F=32.24N[/tex]
A tennis player swings at a ball at a constant speed, taking 0.50 s to rotate her arms and racquet from horizontal to vertical. What acceleration is felt by a small bug at the tip of her racquet if it is 1.3 m from her shoulder
Answer:
the acceleration of the small bug is 12.83 m/s²
Explanation:
Given;
time of motion, t = 0.5 s
radius of the circular path created by his arm, r = 1.3 m
if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰
The centripetal acceleration of the ball is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (\frac{\theta}{t} )^2 r\\\\[/tex]
[tex]a_c = (\frac{90}{360} \times\frac{ 2\pi }{t} )^2r\\\\a_c = (\frac{\pi}{2t} )^2 r\\\\a_c = \frac{\pi^2r}{4t^2} = \frac{\pi^2 \times1.3 }{4\times 0.5^2} = 12.83 \ m/s^2[/tex]
Therefore, the acceleration of the small bug is 12.83 m/s²
An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2900 rpmrpm
Answer:
The torque is 0.31 Nm.
Explanation:
Electrical energy, E = 8400 J
time, t = 1 min
Angular speed, w = 2900 rpm = 303.53 rad/s
efficiency = 2/3 of input power
The toque is given by
[tex]P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm[/tex]
What are the magnitude and direction of a uniform electric field perpendicular to the ground that is able to suspend a particle of mass m = 1.30 g carrying a charge of +6.90 µC in midair, assuming gravity and the electrostatic force are the only forces exerted on the particle?
Answer:
the magnitude and direction of the uniform electric field is 1846.38 N/C and UPWARD respectively
Explanation:
Given the data in the question;
mass m = 1.30 g = 1.30 × 10⁻³ kg
Charge q = +6.90 µC = 6.90 × 10⁻⁶ C
we know that; g = 9.8 m/s
assuming gravity and the electrostatic force are the only forces exerted on the particle, hence the relation is;
F = mg and F = Eq
so
mg = Eq
make E subject of formula
E = mg / q
so we substitute
E = [ (1.30 × 10⁻³) × 9.8 ] / 6.90 × 10⁻⁶
E = 0.01274 / 6.90 × 10⁻⁶
E = 1846.38 N/C
Since the charge is positive ( + ), { direction is Upward }
Therefore, the magnitude and direction of the uniform electric field is 1846.38 N/C and UPWARD respectively
A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80 mA is maintained in the loop. What is the magnetic moment of the loop? (Enter the magnitude.)
Answer:
[tex]\mu = 3.36\times 10^{-3}\ A-m^2[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 0.55 T
The radus of the loop, r = 43 cm = 0.43 m
The current in the loop, I = 5.8 mA = 0.0058 A
We need to find the magnetic moment of the loop. It is given by the relation as follows :
[tex]\mu = AI\\\\\mu=\pi r^2\times I[/tex]
Put all the values,
[tex]\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2[/tex]
So, the magnetic moment of the loop is equal to[tex]3.36\times 10^{-3}\ A-m^2[/tex].
heat travel through vacuum by
a. conduction. b.convention
c. radiation. d. both a&b
Answer:
Option C
Explanation:
C. Radiation.....
Answer:
heat travel through vacuum by radiation
If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of vx.
Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75 cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.
Answer:
V = 8.34m/s
Explanation:
Given that
1/2ke^2 = 1/2mv^2 ......1
Where e = 3.75cm = (3.75/100)m
e = 0.0375m
K = 500 N/m
m = 10g = 10/1000
= 0.01kg
Substitute the values into equation 1
0.5×500×(0.0375)^2 = 0.5×0.01×v^2
250×0.001395 = 0.005v^2
0.348 = 0.005v^2
v^2 = 0.348/0.005
v^2 = 69.6
V = √69.6
V = 8.34m/s
The ball launches at the speed of V = 8.34m/s
A toy plane weighing 1.5 kg is flying in the air at 8 m/s. The player
turns down its speed to 6m/s. What is the work done on it?
Use the work-energy theorem. The total work done on the plane is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (1.5 kg) (6 m/s)² - 1/2 (1.6 kg) (8 m/s)²
W = -21 J
(Note that the work must be negative since the "braking" force opposes the direction in which the plane is moving.)
Un gas a una temperatura de 38°C, tienen un volumen de 21 Lts Litros. ¿Qué volumen tendrá si la temperatura sube a 67°C?
English Translation :
A gas at a temperature of 38 °C, have a volume of 21 Lts Liters. What volume will it have if the temperature rises to 67°C?
Solution :
Since, there is no information about pressure, let us assume it is constant.
So, by ideal gas equation at constant pressure :
[tex]V_1 T_2 = V_2 T_1[/tex]
Putting given volume and temperature ( in Kelvin ) in above equation, we get :
[tex]21 \times ( 67 + 273 ) = V_2 \times ( 38 + 273 )\\\\V_2 = \dfrac{21 \times (67+273) }{(38+273)}\\\\V_2 = 22.96 \ L[/tex]
Hence, this is the required solution.
Three resistors 4ohms 6ohms8ohms are connected in series and are connected to cell of EMF 60volt and negilible resistance calculate the current in circuit , potential difference in resistors ,total terminal potential difference in circuit and lost volt
Explanation:
Given that,
Three resistors 4ohms 6ohms 8ohms are connected in series and are connected to cell of EMF 60volt.
The equivalent resistance in series combination is given by :
R = R₁ + R₂ +R₃
Put all the values,
R = 4 + 6 + 8
R = 18 ohms
Let I is the current in circuit. So,
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{60}{18}\\\\I=3.33\ A[/tex]
Potential difference in 4 ohms,
[tex]V_1=IR_1\\\\V_1=3.33\times 4\\\\=13.32\ V[/tex]
Potential difference in 6 ohms,
[tex]V_2=IR_2\\\\V_2=3.33\times 6\\\\=19.98\ V[/tex]
Potential difference in 8 ohms,
[tex]V_3=IR_3\\\\V_3=3.33\times 8\\\\=26.64\ V[/tex]
Terminal potential difference in circuit is :
V = IR
Put all the values,
V = 3.33 × 18
V = 59.94 volts
Hence, this is the required solution.
1.Anna Traveled 500 m to their house in 600 s. What is the speed of Anna?
A. 0.83 m/s B.2 m/s C.2.7m/s D. 3 m/s
2.What is the velocity of an object that traveled a distance of 100m due east in 10s?
A.10 m/s² B. 10m/s due east C. 10 m east D.10 m/s
3.An object traveled a distance of 50 m in 10 s,What is the speed is the object?
A. 5m/s B.5m/s² C.5m/s2 D. 5m/s due west
1=A.0.83m/s
2=10m/s
3=5m/s
find the equivalent resistance.
Answer:
Explanation:
Note the connection between D-L-H shorts out the resistors on the right. Those 4 resistors can be ignored.
The connection between D-L-K means the 2 resistors between C-D and C-K are in parallel. The equivalent resistance of 2 6-ohm resistors in parallel is 3 ohms.
Similarly the connection between K-L-H means the 2 resistors between J-K and J-H are in parallel. The equivalent resistance of 2 6-ohm resistors in parallel is 3 ohms.
Adding the resistors in series together:
A-C-L-J-D = 6 + 3 + 3 + 6
= 18 ohms
A singly charged ion (q=−1.6×10−19) makes 7.0 rev in a 45 mT magnetic field in 1.29 ms. The mass of the ion in kg is
Answer:
[tex]m=1.47\times 10^{-24}\ Kg[/tex]
Explanation:
Given that,
Charge, [tex]q=1.6\times 10^{-19}\ C[/tex]
Revolution = 7 rev
magnetic field, B = 45 mT
Time, t = 1.29 ms
We need to find the mass of the ion. Let m be the mass. The formula for the mass in terms of time period is given by :
[tex]m=\dfrac{qBT}{2\pi}\\\\m=\dfrac{1.6\times 10^{-19}\times 45\times 10^{-3}\times 1.29\times 10^{-3}}{2\pi}\\\\m=1.47\times 10^{-24}\ Kg[/tex]
So, the mass of the ion is equal to [tex]1.47\times 10^{-24}\ Kg[/tex].
You need to produce a set of cylindrical copper wire 3.5 m long that will have a
resistance of 0.125 Ω each. What will be the mass of each of these wires?
(ρ = 1.72X10-8 Ωm, density of copper = 8.9X103 kg/m3)
Solution :
We know, resistance is given by :
[tex]R = \dfrac{\rho l}{A}[/tex]
[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]
Now, we know mass of wire is given by :
[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]
Hence, this is the required solution.
Given:
Length of wire, l = 3.5 mResistance, R = 0.125 ΩThe resistance will be:
→ [tex]R = \frac{\rho l}{A}[/tex]
or,
→ [tex]A = \frac{\rho l}{R}[/tex]
By substituting the values, we get
[tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]
[tex]= 4.816\times 10^{-7} \ m^2[/tex]
hence,
The mass will be:
→ [tex]Mass = Density\times Volume[/tex]
[tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]
[tex]= 0.01500 \ kg[/tex]
[tex]= 15.00 \ g[/tex]
Thus the above answer is right.
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