Answer:
The total momentum of the cars before the collision is 61,000 kg.m/s
The total momentum of the cars after the collision is 61,000 kg.m/s
The velocity of the cars after the collision is 27.727 m/s
Explanation:
Given;
mass of the first car, m₁ = 1000 kg
initial velocity of the car, u₁ = 25 m/s
mass of the second car, m₂ = 1200 kg
initial velocity of the second car, u₂ = 30 m/s
The common velocity of the cars after collision = v
The total momentum of the cars before collision is calculated as;
P₁ = m₁u₁ + m₂u₂
P₁ = (1000 x 25) + (1200 x 30)
P₁ = 61,000 kg.m/s
The total momentum of the cars after collision is calculated as;
P₂ = m₁v + m₂v
where;
v is the common velocities of the cars after collision since they stick together.
P₂ = v(m₁ + m₂)
To determine "v" apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(1000 x 25) + (1200 x 30) = v(1000 + 1200)
61,000 = 2,200v
v = 61,000/2,200
v = 27.727 m/s
The total momentum after collsion = v(m₁ + m₂)
= 27.727(1000 + 1200)
= 61,000 kg.m/s
Thus, momentum before and after collsion are equal.
Question 2 of 32
A water-skier with a mass of 68 kg is pulled with a constant force of 980 N by
a speedboat. A wave launches him in such a way that he is temporarily
airbome while still being pulled by the boat, as shown in the image below.
Assuming that air resistance can be ignored, what is the vertical acceleration
that the water-skier experiences on his return to the water surface? (Recall
that g = 9.8 m/s2)
Rope Force
ODON
Weight
O A. - 18.1 m/s2
OB. - 15.6 m/s2
O C. -11.2 m/s2
OD. -9.8 m/s2
Answer:
OD. -9.8 m/s2
Explanation:
The only force vertical force that is acting on the skier is gravity and since its pulling him back it's a negative force down the y axis.
a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s. the stone clears the cliff edge on the way down and falls all the way to the ground. what is the maximum height of the stone above the ground
Answer:
h = 90.10 m
Explanation:
Given that,
A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m
The initial speed of the stone, u = 10 m/s
The path followed by the projectile is given by :
[tex]h(t)=-4.9t^2+10t+85[/tex] ....(1)
For maximum height,
Put dh/dt = 0
So,
[tex]\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s[/tex]
Put the value of t in equation (1).
[tex]h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m[/tex]
So, the maximum height of the stone is equal to 90.10 m.
Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?
a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.
Answer:
Option (a).
Explanation:
Let the angular velocity is w.
The centripetal acceleration is given by
[tex]a = r w^2[/tex]
where, r is the distance between the axle and the spoke.
So, more is the distance more is the centripetal acceleration.
(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.
The statement is true.
(b) The direction of centripetal acceleration is always towards the center, so the statement is false.
(c) It is false.
(d) It is false.
Option (a) is correct.
Help please help please
Answer:
No. D is the right answer
I’ve been stuck please help !!
Answer:
The slope of the position time graph gives the velocity.
Explanation:
The slope of the position time graph gives the value of velocity.
In first graph,
The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts. and more than the second part, so the initial velocity is more than the final velocity.
In the second graph,
The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
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A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
sometimes balance point may not be obtained on the potentiometer wire why
A compact car has a maximum acceleration of 3.0 m/s2 when it carries only the driver and has a total mass of 1300 kg. What is its maximum acceleration after picking up four passengers and their luggage, adding an additional 400 kg of mass?
Answer:
[tex]a_2=3.88m/s^2[/tex]
Explanation:
From the Question we are told that:
Initial Mass [tex]m_1=1300kg[/tex]
Final mass [tex]m_2=1300+400=>1700kg[/tex]
[tex]a_1=3.0m/s^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=m_1a_1[/tex]
[tex]F=1300*5[/tex]
[tex]F=6500N[/tex]
Generally the equation for Final acceleration is mathematically given by
[tex]F'=m_2*a_2[/tex]
[tex]a_2=\frac{6500}{1700}[/tex]
[tex]a_2=3.88m/s^2[/tex]
A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?
Answer:
3.73 * 10^16 photons/sec
Explanation:
power supply = 3.0 V
Emits 440 nm blue light
current in LED = 11 mA
efficiency of LED = 51%
Calculate the number of photons per second the LED will emit
first step : calculate the energy of the Photon
E = hc / λ
=( 6.62 * 10^-34 * 3 * 10^8 ) / 440 * 10^-9
= 0.0451 * 10^-17 J
Next :
Number of Photon =( power supply * efficiency * current ) / energy of photon
= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17
= 3.73 * 10^16 photons/sec
A toddler weighs 10 kg and raises herself onto tiptoe (on both feet). Her feet are 8 cm long with each ankle joint being located 4.5 cm from the point at which her feet contact the floor. While standing on tip toe:
(a) what is the upward normal force exerted by the floor at the point at which one of the toddler's feet contacts the floor?
(b) what is the tension force in one of her Achilles tendons? (c) what is the downward force exerted on one of the toddler's
ankle joints?
Answer:
a.49 n
b. 63 n
c. 112 n
Explanation:
a.10 times 9.8 from gravity/2 = 49 n
b. 49n times 4.5/8-4.5 = 63 n
c 49n + 63 n = 112 n
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.
Required:
How much thermal energy is added to the air by the drag force?
Answer:
by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Explanation:
Given the data in the question;
drag coefficient of Cd = 1.4
speed v = 6.0 m/s
One model expands to a square 1.8 mm on a side
Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²
distance travelled s = 240 mm = 0.24 m
we know that; density of air e = 1.225 kg/m³
Now,
Dragging force F[tex]_D[/tex] = ( Cd × e × v² × A ) / 2
thermal energy = F[tex]_D[/tex] × s
so
thermal energy = ( 1.4 × 1.225 × (6)² × (3.24 × 10⁻⁶) × 0.24 ) / 2
thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2
thermal energy = 2.4004512 × 10⁻⁵ J
Therefore, by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a. 50km/hr.
b. 10km/hr
Explanation:
Average speed, which is calculated by dividing the total distance travelled by the time interval as follows:
Average speed = total distance travelled ÷ time
Average velocity is calculated by dividing the total displacement by the time interval as follows:
Average velocity = change in displacement (∆x) ÷ time (t)
According to this question, a motorist travels due North at 90 km/h for 2 hours. She then changes direction and travels West at 60 km/for 1 hour.
Total distance of this journey is 90 + 60 = 150
Total time taken = 1 + 2 = 3hours
Average speed = 150/3
= 50km/hr.
b.) Average velocity = x2 - x1/t
Average velocity = 90 - 60/3
= 30/3
= 10km/hr
the magnitude of the electrical force acting between a +2.4x10-8c charge and 1+1.8x10-6 charge that are separated by 1.008m is
Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
suppose a 1 square meter panel of colar cells has an efficiency of 20% and recieves the equivlent of 6 hours of direct sunlight per day. What average power, in watts, does the panel produce
Answer:
The average power per day is 1008 kW.
Explanation:
Solar constant = 1.4 kW/m2
efficiency = 20 %
area, a = 1 square meter
time = 6 hours
Energy falling on the panel in 6 hours = 1.4 x 6 x 3600 kJ
The output is
= 20 % of 1.4 x 6 x 3600
= 0.2 x 1.4 x 6 x 3600
= 6048 kJ
Average power per day is
= 6048/6 = 1008 kW
Find the weight of a man whose mass is 40 kg on earth.
(also
write complete data plus proper formula).
I am sure it help you with that much ☺️
Explanation:
pleasae give me some thanks please good morning sister
Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be 3.7 m/s. An archerfish spots an insect sitting 18 cm above the waterline and a horizontal distance of 28 cm away. The fish aims its stream at an angle of 39° from the waterline.
Required:
Determine the height above the waterline that the stream reaches at the horizontal position of the insect.
Answer:
The fish gobbles the mosquito at height 18 cm.
Explanation:
Initial velocity, u = 3.7 m/s
horizontal distance, d = 28 cm
Angle, A = 39 degree
Let the time is t.
Horizontal distance = horizontal velocity x time
d = u cos A x t
0.28 = 3.7 cos 39 x t
t = 0.097 s
Let the height is h.
Use the second equation of motion
[tex]h =u t-0.5 gt^2\\\\h= u sin A t - 0.5 gt^2\\\\h= 3.7 sin 39 \times 0.097 - 0.5\times 9.8\times 0.097\times0.097\\\\h =0.226 -0.046 \\\\h=0.18 m=18 cm[/tex]
A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes 0.9 s for the wave to travel the 26 m to the opposite tower. If one meter of the rope has a mass of 0.28 kg, find the tension in the tightrope.
Answer:
the tension in the tightrope is 233.68 N
Explanation:
Given the data in the question;
Time taken to reach the opposite tower t = 0.9 s
Distance between the two towers S = 26 m
mass per one meter length = 0.28 kg
First we calculate the velocity;
Velocity V = Distance / time
we substitute
Velocity V = 26 m / 0.9 s
Velocity V = 28.889 m/s
We know that Velocity V can also be expressed as;
V = √( T / m )
we make T the subject of formula
V² = T / m
T = mV²
we substitute
T = 0.28 × ( 28.889 )²
T = 233.68 N
Therefore, the tension in the tightrope is 233.68 N
A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the south bank pointed in a direction 26o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground
Answer:
(a) The speed is 7.96 m/s
(b) The direction is 76 degree from positive X axis in counter clockwise direction.
Explanation:
Width of river = 280 m
speed of river, vR = 4.7 m/s towards east
speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north
[tex]vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j[/tex]
(a) The velocity of boat with respect to ground is
[tex]\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s[/tex]
(b) The direction is given by
[tex]tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o[/tex]
what are 2 ways that we can express to show our connection to our culture
Answer:
Food, clothes, language, and belief
2 Lights slows down when it enters water from air.
a What happens to its speed?
b What happens to its wavelength?
c What happens to its frequency?
semiconductor have negative temperature coefficient of resistance why
Answer:
As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.
Explanation:
Which element makes up most of the Sun?
A. Sodium
B. Carbon
C. Lithium
D. Hydrogen
Answer:
D. Hydrogen
Explanation:
The sun is a big ball of gas and plasma. Most of the gas — 91 percent — is hydrogen.
Answer:
D. Hydrogen
Explanation:
Hydrogen makes up most of the Sun. It is nearly 91 percent.
violations in a responsible manner in a democratic society?
Activity 5:
Ending
From your findings, what conclusions and recommendations can you make on the
issue of human rights violations to:
5.1
Government
Answer:
Kindly check explanation
Explanation:
The trampling and violation of human rights individuals, groups and corporate organizations is really alarming ad as such, the government who are charged to protect the right and interest of its citizen. In other to curtail the trending issues of human right violation, it is imperative if sensitization programmes could be organiz d in other to keep people informed of the various ways in which people's right may be trampled upon. With these education, the ignorance can be expunged leaving only those who genuinely decides to relate and threaten the right of his fellow country person.
The laws on human right violation should be reviewed and capital punishment metted on violators in other to send a strong warning to those who still nurture the intention.
You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s.
1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.
Answer:
Explanation:
Given that:
the putty initial speed (u) = 9.10 m/s
distance (s) between hand and the ceiling = 3.50 m
the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.
v² - u² = 2as
Since the putty is moving in a vertical motion(i.e. in an upward direction)
v² - u² = -2gs
v² = u² - 2gs
[tex]v = \sqrt{u^2 - 2gs}[/tex]
[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]
[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]
[tex]v = \sqrt{82.81 -68.6}[/tex]
[tex]v = \sqrt{14.21}[/tex]
v = 3.77 m/s
2.
The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:
v = u + at
In an upward direction
v = u - gt
making time t the subject;
v - u = -gt
[tex]t = \dfrac{v-u}{-g}[/tex]
[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]
[tex]t = \dfrac{-5.33}{-9.8}[/tex]
t = 0.54s
does net force stay the same when a massless pulley is replaced by a pulley with mass
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of 1.1 m/s, while Ender recoils with a velocity of -4.3 m/s. Determine the ratio Bonzo/mEnder of the masses of these two enemies.
Answer:
the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
Explanation:
Given the data in the question;
Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s
Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s
the ratio Bonzo/mEnder of the masses of these two enemies = ?
Now, using the law of conservation of momentum.
momentum of both Bonzo and Ender are conserved
so
Initial momentum = final momentum
we have
0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]
[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex] ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]
we substitute
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -4.3 m/s / 1.1 m/s ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -3.9090 ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = 3.91
Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
Suppose that 2 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 35 cm
Answer:
0.035 J
Explanation:
Applying,
W = ke²/2.............. Equation 1
Where W = workdone by the stretching the spring, k = spring constant, e = extension.
make k the subject of the equation
k = 2W/e²............... Equation 2
From the question
Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m
Substitute these values into equation 2
k = (2×2)/(0.15²)
k = 177.78 N/m
Hence, work need to stretch the spring from 33 cm to 35 cm
therefore,
e = 35-33 = 2 cm = 0.02 m
Substitute into equation 1
W = 177.78(0.02²)/2
W = 0.035 J
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
Answer: 1 cal is 4.186 J, 1 kcal = 4186 J A : 1014 m , B 200 m
Explanation: A) Work done by climber is change in potential energy.
W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.
Solve h = 160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m
B Energy is only 20 % : Then h = 0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.
Actually, muscles also produce heat from most of the energy provided by food.
