Answer:
W = 354.9 J
Explanation:
Given that,
The mass of a bale of hay, m = 23 kg
The displacement, d = 3.9 m
The horizontal force exerted on the hay, F = 91 N
We need to find the work done. We know that,
We know that,
Work done, W = Fd
So,
W = 91 N × 3.9 m
W = 354.9 J
So, the required work done is 354.9 J.
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.
A canoe has a velocity of 0.330 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.540 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
Answer:
The velocity of the canoe relative to the river is 0.385 m/s, S37.26⁰W
Explanation:
Given;
velocity of the canoe relative to the earth, [tex]V_{r/e} = 0.33 \ m/s[/tex]
velocity of the river relative to the earth, [tex]V_{r/e} = 0.54 \ m/s[/tex]
The velocity of the canoe relative to the river is calculated as;
[tex]V_{(c/r)x} = V_{(c/e)x}- V_{(r/e)x} \ \ ----(1)\\\\V_{(c/r)y} = V_{(c/e)y}- V_{(r/e)y} \ \ ----(2)[/tex]
The x - component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)x} = 0.33 \times cos \ 45^0\\\\V_{(c/e)x} = 0.2333 \ m/s[/tex]
The y-component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)y} = 0.33 \times sin \ 45^0\\\\V_{(c/e)y} = 0.2333 \ m/s[/tex]
Note: velocity of the river relative to the earth has only x-component = 0.54 m/s
Apply equation (1) and (2) to calculate the velocity of the canoe relative to the river;
[tex]V_{(c/r)}x = 0.2333 - 0.54 = -0.3067 \ m/s\\\\V_{(c/r)}y = 0.2333 - 0 = 0.2333 \ m/s\\\\The \ resultant \ velocity;\\\\V_{c/r} = \sqrt{(-0.3067)^2 + (0.2333)^2} \\\\V_{c/r} = 0.385 \ ms/\\\\The \ direction:\\\\\theta = tan^{-1} (\frac{0.2333}{0.3067} ) = 37.26^0 \ south \ west \ of \ the \ river[/tex]
The decibel level of a jackhammer is 125 dB relative to the threshold of hearing. Determine the decibel level if three jackhammers operate side by side.
Answer:
130 dB
Explanation:
The equation for decibel level is given by:
[tex]D=10log(\frac{I}{I_n} )\\\\Where\ D\ is\ the \ decibel\ level\ in\ dB, I\ is\ the\ intensity\ in \ W/m^2, \\I_n\ is\ threshold\ intensity\ to\ the\ human\ ear=1*10^{-12}W/m^2\\\\Given\ that\ D=125dB, hence:\\\\125=10log(\frac{I}{1*10^{-12}} )\\\\12.5=log(\frac{I}{1*10^{-12}} )\\\\I=3.2\ W/m^2[/tex]
The intensity for 1 jack hammer is 3.2 W/m², therefore for 3 jack hammers, the intensity = 3 * 3.2 = 9.6 W/m²
[tex]D=10\ log(\frac{I}{I_n} )\\\\D=10*log(\frac{9.6}{1*10^{-12}} )\\\\D=130\ dB[/tex]
A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surface of the water directly below the bird. Does the bird appear to the diver to be closer to or farther from the surface than the actual bird
Answer:
The bird appears farther
Explanation:
This is because as the light from the bird travels into the water which has a higher refractive index than air, light rays from the kingfisher bird bend towards the normal at the water surface and thus enter the eye of the scuba diver. Now, if we project the light rays from the eyes of the scuba diver into the air, we see that they appear to come from a point farther than that of the actual kingfisher bird perched on the branch.
So, the bird appears to the diver to be farther from the surface than the actual bird
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
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.
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
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.
Help please help please
Answer:
No. D is the right answer
44.7
When Xavier places his hands near a light bulb, he notices that certain areas around the light bulb are warmer than
others. Which best explains this?
The areas to the sides of the light bulb are warmest because of conduction,
O The areas to the sides of the light bulb are warmest because of convection,
The area directly above the light bulb is warmest because of conduction,
The area directly above the light bulb is warmest because of convection.
Save and Exit
Submit
Mark this and retum
Nex
Answer:
The area directly above the light bulb is warmest because of convection.
Explanation:
if all the sides of the bulb are equally close to the light source inside the bulb, all area of the bulb would be equally heated by conduction. however, convection heating mainly heats up the surface above the light source. in convection heating, the air above the surface of the light source get heated by the light source and expands, casuing it to be less dense and rise to the top of the bulb. colder denser air at the top of the bulb sink to the light source adn gain heat and expands, becoming less dense. this process repeats and the surface above the light source becomes the warmest due to convection heating
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]
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.
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.
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
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 teacher performs a demonstration to show the properties of an unknown substance. The teacher cuts off a piece of gray shiny substance and drops it in water. The substance floats and reacts violently with the water. Based on this Information, what type of element the unknown substance?
A. metalloid
B. metal
C. nonmental
D. halogen
Answer:
nonmental
Explanation:
Option B is correct. The substance floats and reacts violently with the water. Based on this Information, metal is the unknown substance.
What is metal?A metal, a glossy substance that transmits electricity and heat reasonably efficiently when newly manufactured, polished, or broken.
Metals are either malleable or ductile. Metals can be chemical elements like iron.
Metal is substance floats and reacts violently with the water. When particles collide, energy is transferred. A change in temperature can be used to identify this.
The substance floats and reacts violently with the water. Based on this Information, metal is the unknown substance.
Hence, option B is correct.
To learn more about metal, refer to the link;
https://brainly.com/question/18153051
#SPJ2
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
what are 2 ways that we can express to show our connection to our culture
Answer:
Food, clothes, language, and belief
sometimes balance point may not be obtained on the potentiometer wire why
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
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20 points and brainliest‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
parallel (at 0°) to a magnetic field of
0.0579 T. What is the magnetic force
on the charge?
Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
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:
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?
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.
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
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]
List the 5 theoretical perspectives that underlie much of the research on human development. Also, name an individual strongly associated with each perspective.
Answer:
25
Explanation:
5 theroical name indvivdual perspective asssssoitive each persp
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
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]
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.
