Answer:
43.2
because to convert from m/sec to kmph we need to multiply by 3600/1000
Which of the following statements are true?
a. By convention, the direction of a current is taken to be the direction of flow for negative charges.
b. When an electric field is applied to a conductor, the free electrons move only in the direction opposite the applied electric field.
c. Current is the total amount of charge that passes through a conductor's full cross section at any point per unit of time.
d. In order to maintain a steady flow of current in a conductor, a steady force must be maintained on the mobile charges.
e. In a circuit, current is delivered by the positive terminal of a battery, and it is used up by the time it returns to the negative terminal of the battery.
Answer:
A, C, D and E are the true statement about current flowing.
During a phase change the temperature of a substance remains constant this is because during a phase heat changes the ____ energy of particles in a substance without changing their ____ energy
Answer:
Explanation:
individual and then net
hope that helps I could be wrong about this one though
4. Taylor Swift pushes Kanye West,
headfirst fearless, off a high dive with and
initial speed of 6m/s. If the platform is 20m
high, what is Kanye's impact velocity with
the water? (velocity and angle)
Basketball player Darrell Griffith is on record as
attaining a standing vertical jump of 1.2 m (4 ft).
(This means that he moved upward by 1.2 m after
his feet left the floor.) Griffith weighed 890 N (200
lb). g=9.8 m/s2
1- What is his speed as he leaves the floor?
2- if the time of the part of the jump before his feet left the floor was 0.300s, what was the magnitude of his average acceleration while he was pushing against the floor?
Explanation:
1.
We use the equation
h = [tex]\frac{gt^2}{2}[/tex], where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
[tex]\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }[/tex]
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
Take 47 points. Hurry Someone save me just look the picture
Answer:
suppose buying pizza for a work party leads to positive morale and to the work being done in half the time. Pizza is the independent variable, Work speed is the dependent variable, The mediator, the middle man without which there would be no connection, is positive morale. Denaturation, in biology, the process of modifying the molecular structure of a protein. Denaturation involves the breaking of many of the weak linkages, or bonds (e.g., hydrogen bonds), within a protein molecule that is responsible for the highly ordered structure of the protein in its natural (native) state.
Explanation:
hope this helps have a good rest of your day :) ❤
Explanation:
HW1. For example, suppose buying pizza for a work party leads to positive morale and to the work being done in half the time. Pizza is the independent variable, Work speed is the dependent variable, The mediator, the middle man without which there would be no connection, is positive morale.
HW2. The word denature means to render food unpleasant or dangerous to consume, it is denatured by adding a substance known as a denaturant. Aversive agents—primarily bitterants and pungent agents—are used to produce an unpleasant flavor.
Hope It Helps U
Please Mark As Brainl
You drive on Interstate 10 from San Antonio to Houston, half the time at 75 km/h and the other half at 106 km/h. On the way back you travel half the distance at 75 km/h and the other half at 106 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip
Answer:
Explanation:
a ) from San Antonio to Houston let distance be d km .
Average speed = total distance / total time
time = distance / speed
Total time = (d / 2 x 75 ) +( d / 2 x 106 )
= .0067 d + .0047 d
= .0114 d
Average speed = d / .0114 d = 87.72 km /h
b ) from Houston back to San Antonio
Total time = (d / 2 x 106 ) +( d / 2 x 75 )
= .0047 d + .0067 d
= .0114 d
Average speed = d / .0114 d = 87.72 km /h
c )
For entire trip :
total distance = 2d
total time = 2 x .0114 d
Average speed = 2 d / 2 x .0114 d
= 87.72 km /h .
If vec A and vec B are vectors and vec B = -vec A , which of the following is true ?
a) The magnitude of is equal to the negative of the vec B magnitude of vec A
b) Ā and vec B are perpendicular .
c) The direction angle of is equal to the direction angle of vec B A plus 180 degrees
d ) vec A + vec B =2 vec A .
Answer:
If A + B = C for the vector equation then substituting for B gives
A + (-A) = C = 0
The only thing that can be said about "C" is that it has zero magnitude.
A 20.0 N force is necessary to stretch a spring 0.500 m. What is the spring constant of this spring?
The spring constant : k = 40 N/m
Further explanationGiven
Force = 20 N
The displacement of the spring=x=0.5 m
Required
The spring constant = k
Solution
Hooke's Law
F = k.x
k = F/x
Input the value :
k = 20/0.5
k = 40 N/m
A spring with an unknown spring constant is hung vertically, and a 200 g (0.200 kg) mass is attached to the bottom. If the spring stretches 0.250 m from its resting position to the position at which the hanging mass is in equilibrium, what is the spring constant of this spring?
Answer:
k = 7.84 N/m
Explanation:
We are given;
Mass hanging object; m = 0.2 kg
Extension; Δx = 0.25 m
Now, formula for the force is;
F = k•Δx
Where k is the spring constant
Since we have mass, then F = W = mg = 0.2 × 9.8 = 1.96 N
Thus;
1.96 = k × 0.25
k = 1.96/0.25
k = 7.84 N/m
Two stationary positive point charges, charge 1 of magnitude 3.95 nC and charge 2 of magnitude 1.80 nC, are separated by a distance of 39.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.What is the speed of the electron when it is 10.0 cm from charge 1?
Answer:
the speed of the electron from charge q1 is 7.17×10⁶ m/s
Explanation:
Given the data in the question;
the potential at the center of the two charges will be;
V = k( q1/(d/2) + q2/(d/2)
so we substitute
V = (9×10⁹)( (3.95×10⁻⁹/(0.39/2) + 1.80×10⁻⁹/(0.39/2)
V = 265.4 V
the potential at a distance of 10 cm from the charges will be
V = k( q1/(d1) + q2/(d2)
(d1 = 10cm = 0.1m and d2 = 39cm - 10cm = 29cm = 0.29m )
V' = (9×10⁹)( (3.95×10⁻⁹/0.1 + 1.80×10⁻⁹/0.29
V' = 411.4 V
Now, from the conservation of energy the speed of the electron from charge q1 will be;
E = ( V' - V) qe
1/2mv² = ( V' - V) qe
v² = [( V' - V) qe] / 1/2m
v =√ ([( V' - V) qe] / 1/2m)
v =√ ([2( V' - V) qe] / m)
we substitute
v =√ (2[( 411.4 - 265.4) 1.6×10⁻¹⁹] / 9.1×10³¹)
v = 7.17×10⁶ m/s
Therefore, the speed of the electron from charge q1 is 7.17×10⁶ m/s
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
Answer:
The added mass will mean a longer period of oscillation.
Explanation:
The period of oscillation here is given by the formula;
T = 2π√(m/k)
Where m is mass and k is spring constant
From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.
Thus, the added mass will mean a longer period of oscillation.
A hovercraft is being driven across a lake on a very windy day. The wind exerts a force of 5000 N north on the hovercraft. The propellers driving the hovercraft also exert a force.
What is the net force (to 3.s.f ) on the hovercraft when the force applied to the hovercraft by the propellers is: 10 000 N west
Answer:
The net force on the hovercraft is 11200 N.
Explanation:
Given;
force exerted on the hovercraft by wind, F₁ = 5000 N north
force exerted on the hovercraft by the propeller, F₂ = 10,000 N west
The net force on the hovercraft is calculated as;
[tex]F_{net} = \sqrt{F_1^2 + F_2^2} \\\\F_{net} = \sqrt{5000^2 + 10,000^2} \\\\F_{net} = 11180.34 \ N\\\\F_{net} = 11200 \ N \ (3.s.f)[/tex]
Therefore, the net force on the hovercraft is 11200 N.
5. (Liquids Gases) have particles with enough energy to spread out
throughoutcontainer. *
A)Liquids
B)Gases
Answer:
b gases
Explanation:
this is because gases spread out through out the entire container because they have no definite shape and are always moving.
When an object falls, its:
A. PE increases and KE decreases.
B. PE does not change.
C. PE and KE both increase.
D. PE decreases and kE increases
Answer:
Option D. is correct.
Explanation:
The object's mechanical energy refers to the sum of the potential and kinetic energies of the object. When an object falls, its potential energy (PE) decreases, and its kinetic energy (KE) increases. The increase in kinetic energy is exactly equal to the decrease in potential energy.
Option D. is correct.
Physical science-current can be increased by...
Option 3.) Increasing the voltage across the wire.
I know that the other answers are incorrect because, for one thing, the more resistance in a substance, the less flow of the current there is. Also, using a longer wire doesn't change anything, it just makes a electrical current go on longer. Lastly decreasing the voltage would make the current decrease in the atoms that flow through it to power an object.
-R3TR0 Z3R0
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to volume 1450 cm3. If the expansion is isothermal, what are (a) the final pressure and (b) the work done by the gas
Answer:
a) The final pressure is 1.68 atm.
b) The work done by the gas is 305.3 J.
Explanation:
a) The final pressure of an isothermal expansion is given by:
[tex] T = \frac{PV}{nR} [/tex]
[tex] T_{i} = T_{f} [/tex]
[tex] \frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR} [/tex]
Where:
[tex]P_{i}[/tex]: is the initial pressure = 5.79 atm
[tex]P_{f}[/tex]: is the final pressure =?
[tex]V_{i}[/tex]: is the initial volume = 420 cm³
[tex]V_{f}[/tex]: is the final volume = 1450 cm³
n: is the number of moles of the gas
R: is the gas constant
[tex] P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm [/tex]
Hence, the final pressure is 1.68 atm.
b) The work done by the isothermal expansion is:
[tex] W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J [/tex]
Therefore, the work done by the gas is 305.3 J.
I hope it helps you!
What is the total amount of kinetic and potential energy in a system ?
Answer:
Its the sum of the potential energy and the kinetic energy
What type of reaction is occurring when you activate an instant cold pack?
A. Endothermic, because energy is absorbed
B. Exothermic, because energy is released
O C. Endothermic, because energy is released
D. Exothermic, because energy is absorbed
Answer:
Endothermic, because energy is absorbed (A)
Explanation:
The reaction that take place in the instant cold pack causes the surroundings around it, including the bag that contains it. When the bag loses heat to the cold pack, the cold pack absorbs the heat, thereby causing the environment (the bag) to be cold.
Two people are playing tug-of-war. Due to their choice of footwear, theircoefficient of static friction is different. Participant 1 has a mass of 60 kg, acoefficient of static friction of 2.0, and can pull with a maximum force of1000 N. Participant 2 has a mass of 80 kg and a coefficient of staticfriction of 1.2, and can pull with a maximum force of 1200 N. Who wins?
Answer:
Participant 1 wins
Explanation:
Coefficient of static, μ = F/N, where F is the frictional force and N is the normal force.
The force exerted by each participant is the sum of the frictional force acting on each of them and the maximum force with which each participant pulls on the rope.
Frictional force, F = μ * N
Normal force, N = mass * acceleration due to gravity, g
For Participant 1; μ = 2.0, mass = 60 kg, g = 9.8 m/s²
Frictional force = 2.0 * 60 * 9.8 = 1176 N
Total force = (1176 + 1000) = 2176 N
For Participant 2; μ = 1.2, mass = 80 kg, g = 9.8 m/s²
Frictional force = 1.2 * 80 * 9.8 = 940.8N
Total force = (940.8 + 1200) N = 2140.8 N
From the values obtained above, Participant 1 exerts more force than Participant 2, therefore, Participant 1 wins
a car traveling at 30m/s slows down to a stop 10s. what is the acceleration?
Answer:
20 m/s. have a great day
Answer:
since v decreased by 20m/s in 5 sec, a = -4 m/s^2
assuming the 3 seconds started at t=0,
s = 30t - 2t^2
s(3) = 30(3) - 2(9) = 72m
___is found in fruits and honey. *
1.Maltose
2.Sucrose
3.Fructose
4.Galactose
Answer:
3. Fructose
Explanation:
Fructose is a sugar found naturally in fruits, fruit juices, some vegetables and honey.
it is number 3 (Fructose)
I =p/4pir^2 solving for r
Answer: [tex]r = \sqrt{\frac{p}{4pil}}[/tex]
Explanation:
[tex]l = \frac{p}{4pir^2} \\4pir^2l=p\\r^2 = \frac{p}{4pil} \\r = \sqrt{\frac{p}{4pil}}[/tex]
A(n) 17.4 g bullet is shot into a(n) 5506 g
wooden block standing on a frictionless sur-
face. The block, with the bullet in it, acquires
a speed of 1.61 m/s.
Calculate the speed of the bullet before
striking the block..
Answer in units of m/s.
The initial speed of the bullet = v₁= 511.07 m/s
Further explanationGiven
17.4 g bullet
5506 g wooden
The velocity of the block+bullet :1.61 m/s
Required
The initial speed
Solution
Momentum
m₁v₁+m₂v₂=m₁v₁'+m₂v₂'
v₂=0 ⇒block at rest
v₁'=v₂'=1.61 m/s
the equation becomes :
m₁v₁=(m₁+m₂)v'
17.4v₁=(17.4+5506)1.61
v₁= 511.07 m/s
A 80 N force is needed to slide a 50.0 kg box across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor
Answer:
0.16
Explanation:
Given data
Force F= 80N
Mass m= 50kg
Reaction R= Weight= mg= 50*9.81= 490.5N
We know that
F=UR
Substitute and solve of U
U=F/R
U= 80/490.5
U=0.16
Hence the coefficient of friction is 0.16
1. A plane starts from rest and aceelerates in a
straight line along the ground before take-off. It
moves 600 m in 12 s. Calculate the distance
moved during the twelfth second.
s=600 m
t=12 s
s=0.5*a*t² (initial speed V0=0)
a=(2*s)/t²
a=(2*600)/12²
a≈8.33 m/s²
L= s(t2=12s)-s(t1=11s) -> (distance during the twelfth second)
L=0.5*a*(t2²-t1²)
L=0.5*((2*s)/t²)*(t2²-t1²)
L=0.5*((2*600)/12²)*(12²-11²)
L ≈ 95.83 m
chinese wares are wrapped by piece of paper why
Answer:
chinawares are wrapped by paper while packaging to reduce the chances of the wares breaking when falling
The spectrum of a star shows a set of dark absorption lines equivalent to the absorption lines of the Sun but with one exception: Every line appears at a slightly longer wavelength, shifted toward the red end of the spectrum. What conclusion can be drawn from this observation
Answer:
The shift in the emission spectra of the relative velocity between the star and our planet, is called the relativistic doppler effect
Explanation:
The absorption and emission spectra are measured differently, the emission spectrum is the spectrum emitted by the star and the absorption spectrum is the absorption of these lines by the gases of our atmosphere, this absorption occurs for relatively broad Δλ.
The shift in the emission spectra of the relative velocity between the star and our planet, is called the relativistic doppler effect and has a red shift if the star moves away from us.
[tex]f_o = f_s \sqrt{\frac{1- v/c}{1 + v/c} }[/tex]
A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18.8 N .
(a) Draw two free-body diagrams: one for each block.
(b) What is the acceleration of either block?
(c) Find m.
(d) How does the tension compare to the weight of the hanging block?
Answer and Explanation:
(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ([tex]F_{t}[/tex]), vertical gravitational force ([tex]F_{g}[/tex]) and vertical normal force ([tex]F_{n}[/tex]), due to the surface. Since there is no vertical movement, [tex]F_{g}[/tex] and [tex]F_{n}[/tex] cancels it out. So, for this block, net force is horizontal due to the rope [tex]F_{t}[/tex].
The block of mass m is hanging from the pulley, so there is the force of the rope ([tex]F_{t}[/tex]) and the gravitational force ([tex]F_{g}[/tex]). Both are vertical, because there is no surface "holding" block m.
(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:
[tex]F_{r}=m.a[/tex]
[tex]a=\frac{F_{r}}{m}[/tex]
[tex]a=\frac{18.8}{3.6}[/tex]
a = 5.22
The acceleration of either block is 5.22 m/s².
(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².
Suppose positive referential is going up. To determine mass:
[tex]F_{r}=m.a[/tex]
[tex]F_{t}-F_{g}=m.a[/tex]
[tex]F_{t}-m.g=m.a[/tex]
[tex]18.8-9.8m=5.22m[/tex]
[tex]15.02m=18.8[/tex]
m = 1.25
Block m has 1.25 kg.
(d) Gravitational force is also called weight. So, as described above: [tex]F_{g}=m.g[/tex].
The weight for the hanging block is
[tex]F_{g}=1.25*9.8[/tex]
[tex]F_{g}=[/tex] 12.25 N
Comparing tension and weight:
[tex]\frac{12.25}{18.8}[/tex] ≈ 0.65
We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.
The bat emits a sound wave with a frequency of 25.0 kHz and a wavelength of 0.0136
metres.
Calculate the speed of this sound wave.
Answer:
The calculation that you then need to do is 25000 x 0.0136 = 340. You must not forget the units of speed, which here are metres per second, or m/s. Your final answer is 340 m/s.
A student drove to the university from her home and noted that the odometer reading of her car increased by 17.9 km. The trip took 26.6 min.
a) What was her average speed, in kilometers per hour?
b) IS she returned home by the same path 7 h 30 min after she left, what was her average speed and velocity for the entire trip?
Answer:
See calculation below
Explanation:
Average speed = Distance/Time
a) Given
Distance = 17.9km
Time = 26.6minutes
Time = 26.6/60 hr
Time = 0.443hr
Average speed = 17.9/0.443
Average speed = 40.38km/hr
b) If she returned home by the same path 7 h 30 min after she left, the distance will be the same;
Distance = 17.9km
Time = 7.5hr
Average speed = 17.9/7.5
Average speed = 2.387km/hr
Velocity of the entire trip = 40.38km/hr + 2.387km/hr
Velocity of the entire trip = 42.77km/hr
Hence the velocity of the entire trip is 42.77km/hr