Answer:
v₀ₓ = 24.24 m / s
Explanation:
This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.
Y axis
initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
when it reaches the ground its height is zero and the initial height is y₀=1.2m
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2 y_o/g}[/tex]
t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]
t = 0.495 s
X axis
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 12 / 0.495
v₀ₓ = 24.24 m / s
3. What is electric current?
The flow of moving electrons
electrons that move one time
Answer:
An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. ... In electric circuits the charge carriers are often electrons moving through a wire.
Answer:
The flow of moving electrons
explanation on energy from air pressure light from water pressure
When you release some of the paint from a spray paint can and the can remains at the same temperature, which gas law does this represent?
Answer:
Boyle's law.
Explanation:
Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
[tex] PV = K[/tex]
[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]
Where;
P1 is the original pressure.
P2 is the final pressure.
V1 is the original volume.
V2 is the final volume.
Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.
Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously deposited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks. Assume the dome has a diameter of 25.0 cm and is surrounded by dry air with a "breakdown" electric field of 3.00 10^6 V/m.
Required:
a. What is the maximum potential of the dome?
b. What is the maximum charge on the dome?
Answer:
(a) V = 3.75 x 10^5 V
(b) q = 5.2 x 10^-6 C
Explanation:
Diameter, d = 25 cm
radius, r = 12.5 cm = 0.125 m
Electric field, E = 3 x 10^6 V/m
(a) The maximum potential is given by
[tex]V = E \times r \\\\V = 3\times 10^6\times 0.125\\\\V = 3.75\times10^5 V[/tex]
(b) The charge is given by
[tex]V = \frac{k q}{r}\\\\3.75\times10^5=\frac{9\times10^9\times q}{0.125}\\\\q = 5.2\times 10^{-6} C[/tex]
Which hand position should be avoided in fitness walking?
flexing wrists
relaxing fingers
clenching fists
keeping hands loose
Answer:
The answer should be clenching fists
In which states of matter will a substance have a fixed volume?
O A. Liquid and solid
O B. Solid and gas
O C. Plasma and gas
O D. Liquid and gas
Answer:
A. liquid and solid
Explanation:
A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km
Answer:
[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]
Answer:
108 km/hr or 0.03 km/s
Explanation:
conversion factor for m/s to km/hr is 5/18
conversion factor for m/s to km/s is 1/1000
Could you please explain the step by step process of setting up this problem?
A 10 kg, 4 m long plank of wood is going to be used as a teeter-totter for a brother and sister. The brother has a mass of 30 kg and the sister a mass of 20 kg. If the brother and sister sit at opposite ends of the plank, how far from the brother should the fulcrum be in order for the teeter-totter to be balanced?
A. 1.33 m
B. 1.60 m
C. 1.67 m
D. 2 m
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to the atmosphere. Find the power to the air and how much air should be flowed to the cooling tower (kg/s) if its temperature cannot be increased more than 10oC.
Answer:
- the power to the air is 850 MW
- mass flow rate of the air is 84577.11 kg/s
Explanation:
Given the data in the question;
Net power generated; [tex]W_{net[/tex] = 150 MW
Heat input; [tex]Q_k[/tex] = 1000 MW
Power to air = ?
For closed cycles
Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]
we substitute
Power to air Q₀ = 1000 - 150
Q₀ = 850 MW
Therefore, the power to the air is 850 MW
given that ΔT = 10 °C
mass flow rate of air required will be;
⇒ Q₀ / CpΔT
we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K
we substitute
⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]
⇒ ( 850 × 10³ ) / 10.05
⇒ 84577.11 kg/s
Therefore, mass flow rate of the air is 84577.11 kg/s
A 1500kg car is travelling at v=30m/s. The cars kinetic energy is? *
A) 45000J
B) 1350000J
C) 22500J
D)675000J
show your work please
Hi there!
[tex]\large\boxed{\text{D. 675000J}}[/tex]
Use the following formula to solve:
KE = 1/2mv², where:
KE = kinetic energy
m = mass (kg)
v = velocity (m/s)
Therefore:
KE = 1/2(1500)(30)²
KE = 1/2(1500)(900)
KE = 675000 J
A car changes speed from 27m/s to 5m/s in 50m. The acceleration is: *
A) 7m/s2
B) 7.04m/s2
C) -7.04m/s2
D) 0.22m/s2
show your work please
by using v ^2 = u^2 + 2as we can find "a"
25 = 729 + 2 × a × 50
25 = 729 + 100a
a = - 7.04
so the answer is B
A train mass of 2000kg and speed 35 m/s collides and sticks to an identical train that is initially at rest .After the collision (a) what is the final speed of the entangled system?
(b) what is the kinetitic energy of the system? compare the final kinetic energy to initial kinetic energy?
Answer:
The system would be moving at [tex]17.5\; \rm m \cdot s^{-1}[/tex].
The kinetic energy of this system would be [tex]612500\; \rm J \![/tex] after the collision.
[tex]612500\; \rm J[/tex] (same amount) of kinetic energy would be lost.
Explanation:
The momentum of an object is the product of its mass [tex]m[/tex] and its velocity [tex]v[/tex]. That is: [tex]p = m \cdot v[/tex].
Assume that external forces (e.g., friction) have no effect on this system. The total momentum of this system would stay the same before and after the collision.
Initial momentum of this system:
Moving train: [tex]\begin{aligned}p &= m \cdot v \\ &= 2000\; \rm kg \times 35\; \rm m \cdot s^{-1} \\ &= 70000\; \rm kg \cdot m \cdot s^{-1}\end{aligned}[/tex].Since the other train wasn't moving before the collision, its initial momentum would be [tex]0[/tex].Hence, the momentum of this system would be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] before the collision.
Under the assumptions, the collision would not change the momentum of this system. Hence, the momentum of this system would continue to be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] after the collision.
However, with two identical trains stuck to each other, the mass of this system would be twice that of just one train: [tex]m = 2 \times 2000\; \rm kg[/tex].
Calculate the new velocity of this system:
[tex]\begin{aligned} v &= \frac{p}{m}\\ &= \frac{70000\; \rm kg \cdot m \cdot s^{-1}}{2 \times 2000\; \rm kg} = 17.5\; \rm m\cdot s^{-1}\end{aligned}[/tex].
Calculate the kinetic energy of this system before and after the collision.
Before the collision:
[tex]\begin{aligned}& \text{KE(before)} \\ =\; & \text{KE(moving train)} + \text{KE(stationary train)}\\ =\; & \frac{1}{2} \, m(\text{one train}) \cdot (v(\text{moving train}))^{2} + 0 \\ = \; &\frac{1}{2} \times 2000 \times (35\; \rm m\cdot s^{-1})^{2} \\ = \; & 1225000\; \rm J \end{aligned}[/tex].
After the collision:
[tex]\begin{aligned}& \text{KE(after)} \\ =\; & \frac{1}{2} \, m(\text{two trains}) \cdot v^{2} \\ = \; &\frac{1}{2} \times (2\times 2000\; \rm kg) \times (17.5\; \rm m\cdot s^{-1})^{2} \\ = \; & 612500\; \rm J \end{aligned}[/tex].
Change to the kinetic energy of this system:
[tex]1225000\; \rm J - 612500\; \rm J = 612500\; \rm J[/tex].
A car starts from rest and accelerates uniformly at 3.0 m/s2 toward the north. A second car starts from rest 6.0 s later at the same point and accelerates uniformly at 5.0 m/s2 towar the north. How long after the second car starts does it overtake the irst car?
a. 12 s
b. 19 s
c. 21 s
d. 24 s
what affects our utility
Answer:
Energy Bill fluctuations are inevitable and depend on a variety of different factors. Two of the most important are the current weather your home is experiencing and the current price per Kilowatt Hour (which fluctuates more than you might think).
A basketball player is getting ready to jump, pushing off the ground and accelerating upward.
A) Draw a force identification diagram.
B) Draw a free body diagram.
Answer:
B
Explanation:
B
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
A positive point charge, Q = 4.00 nC, is located a distance of 5.00 cm above the midpoint of the
rod. What will be the electrical force on the point charge?
A scooter is accelerated from rest at the rate of 8m/s
. How long will it take to cover
a distance of 32m?
Explanation:
time=Distance/speed
t=32/8
t=4 seconds
A heat engine with 0.100 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle Isothermal expansion to 5000 cm3 ?
Part A How much work does this engine do per cycle? Express your answer with the appropriate units. sochoric cooling to 200 K -Isothermal compression to 3000 cm3. - Isochoric heating to 800 K Value Units
Part B What is its thermal efficiency? Express your answer with the appropriate units.
Answer:
below
Explanation:
Part A) This engine works per cycle is 254.9 J.
Part B) The thermal efficiency is 23.42%
What is the thermal efficiency?The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.
For isothermal expansion, work done is
W₁ =nRT₁ x ln(V₂/V₁)
W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)
W₁ = 339.8 J =Q₁
For isochoric cooling ,
W₂ =0
Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)
Q₂ = -748.3 J
For isothermal compression,
W₃ =nRT₂ ln (V₄/V₃)
W₃ = 0.1 x 8.314 x 200 x ln(3000/5000)
W₃ = -84.9J
For isochoric heating
W₄ =0
Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)
Q₄ = -748.3 J
Total work done in all the process W = W₁ +W₂ +W₃ +W₄
W =254.9 J
Thus, the work done is 254.9 J
Thermal efficiency = Work done/Heat taken
η = W/ Q₁ +Q₄
η = [254.9 / 339.8 +748.3 ] x 100 %
η = 0.2342 x 100 %
η = 23.42%
Thus, the thermal efficiency is 23.42%
Learn more about thermal efficiency.
https://brainly.com/question/13039990
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What is the torque in ( lbs-ft ) of a man pushing on a wrench with 65 lbs of force 8 unches from the nut / bolt he is trying to turn?
Explanation:
The torque [tex]\tau[/tex] is given by
[tex]\tau=Fd = (65\:\text{lbs})(\frac{8}{12}\:\text{ft}) = 43.3\:\text{lbs-ft}[/tex]
Question 2:
Inclined Plane
A block (M) weighs 25-N, rests on an inclined plane when it is joined by a sting to a support
(S) as shown in the figure' below. Use g=10 N/Kg.
(S)
B
M
List and classify the forces acting on (M).
Représent, without scaling, the forces acting on (M).
Find the mass of (M).
74. If the string were cut, (M) does not slide. Explain this phenomenon.
15. Determine the mass and weight of (M) on moon.
06
Answer:
we need the block
Explanation:
1×2 =4 lest 74 =345
A bag contains lenses with focal lengths 10 cm, 20 cm and 25 cm which are not marked with their focal length. Describe a simple activity to identify the three types lenses
pls give the answer ASAP!!!!!
Explanation:
ehb-pynw-ayo
joi n fast
When rebuilding her car's engine, a physics major must exert 405 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force in newtons between the piston and cylinder
Answer:
[tex]N=675N[/tex]
Explanation:
From the question we are told that:
Force [tex]F=405N[/tex]
Generally the equation for Normal force in this case is is mathematically given by
[tex]F=\mu_s N[/tex]
Where
Static Friction=[tex]\mu_s[/tex]
[tex]\mu_s=0.6[/tex]
Therefore
[tex]N=\frac{F}{\mu_s}[/tex]
[tex]N=\frac{405}{0.6}[/tex]
[tex]N=675N[/tex]
A professional boxer hits his opponent with a 1035 N horizontal blow that lasts 0.175 s. The opponent's total body mass is 120 kg and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.(a) The opponent's final velocity after the blow(b) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.
Answer:
(a) vf = 1.51 m/s
(b) vf = 36.22 m/s
Explanation:
The rate of change of momentum is equal to the force:
[tex]F = \frac{mv_f-mv_i}{t}[/tex]
[tex]Ft = m(v_f-v_i)[/tex]
where,
F = Force = 1035 N
t = time = 0.175 s
vi = initial speed = 0 m /s
vf = final speed = ?
(a)
m = mass of body = 120 kg
Therefore,
[tex](1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\[/tex]
vf = 1.51 m/s
(b)
m = mass of head = 5 kg
Therefore,
[tex](1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\[/tex]
vf = 36.22 m/s
why acceleration independent variable
Answer:
Explanation:Force and acceleration are directly proportional. ... Mass and acceleration are inversely proportional. In this situation, acceleration changes in response to a change of mass, so mass is the independent variable and acceleration is the dependent variable.
Which of the following phenomenon odd called photoelectric effect?
A. High energy electrons impinge on a metallic Annie which emits electrons
B. A high energy photon emits photons as it slows down
C. A metal absorbs a quanta of light and then emits electrons
D. Two electrons are created from a quanta of light.
two electrons are created from a quanta of kight
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Explain how you can get the magnitude of the average electric field between these two points on the paper, and give the value of this field in Newtons/Coulomb. Show your calculations.
Answer:
-30 N/C
Explanation:
Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m
Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V
Since E = -ΔV/Δx
substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -0.30 V/0.01 m
E = -30 V/m
Since 1 V/m = 1 N/C.
E = -30 N/C
So, the average electric field is -30 N/C
Show that the speed with which a projectile leaves the ground is equal to its speed just before it strikes the ground at the end of its journey, assumilng the firing level equals the landing level.
Answer:
Thus, the velocity at the time of strike is same as the velocity at the time of projection.
Explanation:
Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.
At maximum height , the speed is zero and then the projective comes back on the ground.
Use the third equation of motion
[tex]v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}[/tex]
Now let the velocity at the time of strike is v'.
Use third equation of motion, here initial velocity is zero.
[tex]v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}[/tex]
Thus, the velocity at the time of strike is same as the velocity at the time of projection.
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperature of the system 50 C. ( specific heat water= 4200 J/Kg C , specific heat copper= 390 J/Kg C
Answer:
Approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] (assuming that the boiling point of water in this experiment is [tex]100\; \rm ^\circ C\![/tex].)
Explanation:
Latent heat of condensation/evaporation of water: [tex]2260\; \rm J \cdot g^{-1}[/tex].
Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to [tex]\rm J \cdot g^{-1}[/tex].
Specific heat of water: [tex]4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}[/tex].
Specific heat of copper: [tex]0.39\; \rm J \cdot g^{-1}\cdot K^{-1}[/tex].
The temperature of this calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains increased from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]. Calculate the amount of energy that would be absorbed:
[tex]\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}[/tex].
[tex]\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}[/tex].
Hence, it would take an extra [tex]585\; \rm J + 31500\; \rm J = 32085\; \rm J[/tex] of energy to increase the temperature of the calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
Assume that it would take [tex]x[/tex] grams of steam at [tex]100\; \rm ^\circ C[/tex] ensure that the equilibrium temperature of the system is [tex]50\; \rm ^\circ C[/tex].
In other words, [tex]x\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] would need to release [tex]32085\; \rm J[/tex] as it condenses (releases latent heat) and cools down to [tex]50\; \rm ^\circ C[/tex].
Latent heat of condensation from [tex]x\; \rm g[/tex] of steam: [tex]2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J[/tex].
Energy released when that [tex]x\; {\rm g}[/tex] of water from the steam cools down from [tex]100\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]:
[tex]\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}[/tex].
These two parts of energy should add up to [tex]32085\; \rm J[/tex]. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
[tex](2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 13[/tex].
Hence, it would take approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] for the equilibrium temperature of the system to be [tex]50\; \rm ^\circ C[/tex].
When using the process of evaporation to separate a mixture what is left behind to an evaporating dish
A. The mixture does not separate in the entire mixture remains in the dish
B. The liquid evaporates in the solid is left in the dish
C. The mixture does not separate in the entire mixture evaporates
D. None of these
Answer:
B
Explanation:
The liquid evaporates in the solid is left in the dish..
please helpp!
convert 1N into dyne
In the given relation F=ma a stands for write there SI unit
Answer:
a. 1 Newton = 100000 Dyne
b. a represents acceleration.
Explanation:
Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;
1 Newton = 10⁵ Dyne
1 Newton = 100000 Dyne
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Force = mass * acceleration
[tex] F = ma[/tex]
Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.