Given:
The mass of the Darin is m = 75.5 kg
The air resistance is
[tex]F_a=128\text{ N}[/tex]Required: Darin's acceleration.
Explanation:
According to Newton's second law, the downward force will be
[tex]F_g=\text{ mg}[/tex]Here, g = -9.8 m/s^2 is the acceleration due to gravity.
On substituting the values, the downward force will be
[tex]\begin{gathered} F_g=75.5\times(-9.8) \\ =\text{ -739.9 N} \end{gathered}[/tex]The net force will be
[tex]\begin{gathered} F_{net}=\text{ F}_g+F_a \\ =-739.9+128 \\ =-611.9\text{ N} \end{gathered}[/tex]Final Answer: Darin's acceleration is -611.9 N
Isaac Newton in his 1670's lectures on optics and later in his 1704 publicationOpticks, he described light as composed of in his corpuscular hypothesisarguing that the perfectly straight lines of reflection demonstrated this nature.linesparticlesphotonswaves
The perfectly straight lines of reflection demonstrated in the nature is due to particle nature of the light. Therefore, Newton described that the light consists of 'particles' which means second option is correct.
Harry Hand can run 100m in 20s. His daughter, Linda Hand, canrun 50m in 8.5s. Who was faster
The speed is given as;
[tex]v=\frac{d}{t}[/tex]Here, d is the distance covered and t is the time.
Harry Hand covers a distance of 100 m (d_H) in 20 s (t_H). Therefore, the speed of the Harry Hand is,
[tex]v_H=\frac{d_H}{t_H}[/tex]Substituting all known values,
[tex]\begin{gathered} v_H=\frac{100\text{ m}}{20\text{ s}} \\ =5.0\text{ m/s} \end{gathered}[/tex]Now, Linda Hand covers a distance of 50 m (d_L) in 8.5 s (t_L). Therefore, the speed of Linda Hand is given as,
[tex]v_L=\frac{d_L}{t_L}[/tex]Substituting all known values,
[tex]\begin{gathered} v_L=\frac{50\text{ m}}{8.5\text{ s}} \\ \approx5.88\text{ m/s} \end{gathered}[/tex]Since the speed of Linda Hand is greater than Harry Hand (v_L>v_H). Therefore, Linda Hand is faster.
What word am I looking for I have already tried losing
We will have the following:
neutral object becomes charged by losing electrons.
What is the car's velocity between 11h and 15h?
Equation:
df-di/ t f-ti
The velocity of the car between time 11 hr and 15 hr which is shown in the graph would be 4 m/s².
The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.
We are given that,
Displacement of the car = Δx = (20km) - (4km) = 16 km
Time interval of the car = Δt = (15h)- (11h) = 4hours
v = dx/dt
dx = v dt
∫dx = ∫v dt
Δx = v Δt
v = Δx/Δt
Therefore, for get the value of velocity between the given time interval , putting the value in in above equation,
v = 16km/4hours
v = 4 km/hours
Thus, The velocity of the car between time 11 hr and 15 hr will be given as 4km/hours.
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10. A boy of mass 55kg runs at 12m/s and hops on a 15kg skateboard that was at rest. What is thevelocity of the boy on the skateboard afterwards?
M = mass of the boy = 55kg
V = initial velocity of the boy = 12 m/s
m= mass of stationary skateboard = 15kg
v= velocity os stationary sketeboard= 0 m/s
V' = velocity of the boy on the skateboard after collision
Conservation of momentum:
MV + mv = (M + m) V'
Replacing:
55 kg * 12 m/s + 15 kg *0 = (55 kg+ 15 ) V'
Solve for V´'
660 = 70 V´'
660/70 = V'
V'= 9.42 m/s
For an object spinning around a central point, what will happen if its distance from the center is decreased
Answer:
Its a acceleration will increase
Explanation:
The force required to keep an object in a circular motion is given by
[tex]F=\frac{mv^2}{R}[/tex]where v is the radial velocity and R is the radius of the object with mass m.
Now our question is what happens to the above equation as we decrease R?
We can see that as R decreases the quantity mv^2 /R increases (since R is getting smaller ).
Hence, we conclude that F increases. But what if F? it is the centripetal force.
Since centripetal force has increased, so has the quantity v^2 /R (called the acceleration ).
Meaning an increase in centripetal force implies an increase n acceleration.
Since in the answer choices we are not given the option to increase our centripetal force, the next best choice is to choose 'acceleration will increase. '
given q= 10^-6 c, v= (3i + 4j )m/(s) , b=1i tesla find magnetic force
Lorentz magnetic force is the force exerted on a charged particle 'q' which moves with a velocity 'v' through a magnetic field B. The Lorentz forces exerted by moving charges on one another are not equal and opposite.
Given:
Charge, q = [tex]10^{-6}[/tex] C
Velocity, v = (3i + 4j) m/s
Magnetic field, B = 1i T.
To find: Magnetic force, F.
From Lorentz force law, the magnetic force is given by,
F = q × (v × B)
Substituting the given values,
F = [tex]10^{-6}[/tex] × ((3i + 4j) × 1i)
Since i×i = 0, and j×i = k,
F = [tex]10^{-6}[/tex] × 4k
∴ F = 4 × [tex]10^{-6}[/tex] k newtons.
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propose a hypothesis as to how initial speed may affect the range of a projectile
The range, R, of a projectile motion is directly proportional to the horizontal component of the initial velocity as given by the formula,
[tex]R=u_xt[/tex]Where u_x is the horizontal component of the initial velocity and t is the time of flight of the projectile.
The range of a projectile is directly proportional to the square of the initial velocity as given by,
[tex]R=\frac{u^2\sin 2\theta}{g}[/tex]Thus the initial speed increases, the range of the projectile increases.
A snowmobiler travels 65km [37° E of S]. How far east does she travel?
ANSWER:
39.1 km
STEP-BY-STEP EXPLANATION:
To better understand the problem, we make a sketch, like this:
Therefore, we can determine the distance east with the cosine function, like this:
[tex]\begin{gathered} \cos\theta=\frac{\text{ adjacent}}{\text{ hypotenuse}} \\ \\ \theta=53\degree \\ \\ \text{ adjacent =}E \\ \\ \text{ hypotenuse = 65} \\ \\ \text{ We replacing:} \\ \\ \cos53\degree=\frac{E}{65} \\ \\ E=65\cdot\cos53\degree \\ \\ E=39.1\text{ km} \end{gathered}[/tex]The distance to the east is 39.1 km
A locomotive enters a station with an initial velocity of 19 m/s and slows down at a rate of as it goes through. If the station is 175 m long, how fast is it going when the nose leaves the station?
ANSWER
[tex]9m\/s[/tex]EXPLANATION
Parameters given:
Initial velocity, u = 19 m/s
Acceleration, a = -0.8m/s² (the train is slowing down)
Distance traveled, s = 175 m
To find the final velocity of the train, we have to apply one of Newton's equations of motion:
[tex]v^2=u^2+2as[/tex]where v = final velocity
Substituting the given values into the equation, the final velocity of the train as its nose leaves the station is:
[tex]\begin{gathered} v^2=19^2+(2\cdot-0.8\cdot175) \\ v^2=361-280 \\ v^2=81 \\ \Rightarrow v=\sqrt[]{81} \\ v=9m\/s \end{gathered}[/tex]That is the answer.
If an astronaut weighs 148 N on the Moon and 893 N on Earth, then what is his mass on Earth?
_____ kg
The mass on Earth is 91.1 kg
The weight on the Moon is 148 N
The weight on the Earth is 893 N
We need to apply the concept of force.
Weight=massxacceleartion
893=mass of man x acceleration of gravity on earth
893= mass x 9.8
mass = 91.1 kg
Therefore, the mass on earth is 91.1 kg.
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Answer:
the mass on Earth is 91.1 kg
Explanation:
You can hear sound from another room through a door that is slightly open because...the sound refracts as it goes from one room to another.the sound wave reflects off the air in the doorway.the sound diffracts as it goes through the opening.the sound is polarized as it goes through the narrow opening.
when sound wave go from small passages like the edge of a wall or from the slit of open door the sound wave diffracts.
So the 3rd option is correct option.
this is a 2 part question2) Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops byapplying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than,less than, or equal to the stopping distance of driver 2? (b) Choose the best explanation from among thefollowing: 1. Locking up the brakes gives the greatest possible braking force.2. The same tires on thesame road result in the same force of friction.3. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
The maximum static friction between two surfaces is greater than the kinetic friction between them.
If the wheels of a car get locked, the surface of the wheel slides through the floor and kinetic friction acts to stop the car.
If the wheels of the car don't get locked, they may turn fast enough to prevent the surface of the wheel from sliding through the floor and static friction acts on the car.
Since the force acting on the car with its wheel locked is less than the force acting on the car with the turning wheels, then, the stopping distance is greater for driver 1 than for diver 2.
Therefore, the answers are:
a) The stopping distance of driver 1 is greater than the stopping distance of driver 2.
b) The best explanation is:
3. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint, what is the velocity of the ball?
The linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.
What is angular velocity?Angular velocity is the rate of change of angular displacement with respect to time. Mathematically -
ω = dθ/dt
from this we can write -
dθ = ω dt
∫dθ = ω ∫dt
θ₂ - θ₁ = ω(t₂ - t₁)
Δθ = ω Δt
ω = Δθ/Δt
Given is a ball thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. The angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint.
The relation between linear velocity and the angular velocity of a body undergoing circular motion is given by -
v = rω
From this we can write -
ω = 31.3 rad/s
r = 1.45 m
Substituting the values -
v = rω
v = 1.45 x 31.3
v = 45.385 m/s
Therefore, the linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.
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After reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain. In memory, this process is called
In memory, the process in which after reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain is called storage.
In memory there are three phases. They are:
Encoding StorageRetrievalIn the given scenario, the process in which converting the bold, key words to coded representations is encoding process. The process which maintains the coded representations in a network of neurons in your brain is storage. The process of remembering the information stored when needed is retrieval.
Therefore, in memory, this process is called Storage
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Find the direction of the vector A⃗ = (5.1 m )x^ + (-1.5 m )y^.Find the direction of the vector B⃗ = (-1.5 m )x^ + (5.5 m )y^Find the magnitude of the vector A⃗ +B⃗ .Find the direction of the vector A⃗ +B⃗ .
The angle between the components x and y of a vector is given by:
[tex]\theta=\tan ^{-1}(\frac{v_x_{}_{}}{v_y})[/tex]once we know this we need to find in which quadrant the vector lies so we know how to calculate the correct direction.
Vector A lies in the fourth quadrant this means that we need to subtract theta to 360° in order to get the direction of the vector, then we have:
[tex]360-\tan ^{-1}(\frac{1.5}{5.1})=343.61[/tex]Therefore the direction of vector A is 343.61°
Vector B lies in the second quadrant, this means that we need to subtract theta (given by the first equation) to 180° in order to get the direction, then we have:
[tex]180-\tan ^{-1}(\frac{5.5}{1.5})=105.26[/tex]Therefore the direction of vector B is 105.26°
Let's find vector A+B:
[tex]\begin{gathered} \vec{A}+\vec{B}=\langle5.1,-1.5\rangle+\langle-1.5,5.5\rangle \\ =\langle5.1-1.5,-1.5+5.5\rangle \\ =\langle3.6,4\rangle \end{gathered}[/tex]Then we have that:
[tex]\vec{A}+\vec{B}=\langle3.6,4\rangle[/tex]To find its magnitude we have to remember that the magnitude of any vector is given by:
[tex]\lvert\vec{v}\rvert=\sqrt[]{v^2_x+v^2_y}[/tex]Then for vector A+B we have:
[tex]\begin{gathered} \lvert\vec{A}+\vec{B}\rvert=\sqrt[]{(3.6)^2+(4)^2} \\ =5.38 \end{gathered}[/tex]Therefore the magnitude of vector A+B is 5.38 meters.
Vector A+B lies in the first quadrant, then its direction is given by the expression for theta, then we have:
[tex]\tan ^{-1}(\frac{4}{3.6})=48.01[/tex]Therefore the direction of vector A+B is 48.01°
A 1500 kg car traveling at 10 m/s suddenly runs out of gaswhile approaching the valley shown in FIGURE EX10.11. The alertdriver immediately puts the car in neutral so that it will roll.What will be the car’s speed as it coasts into the gas station onthe other side of the valley?
If the car does not brake, its mechanical energy will be conserved along all its trajectory.
Then, the sum of the kinetic energy (K) of the car and its gravitational potential energy (U) will be the same for any two points in the trajectory:
[tex]K_1+U_1=K_2+U_2[/tex]The kinetic energy of an object with mass m and speed v is:
[tex]K=\frac{1}{2}mv^2[/tex]The gravitational potential energy of an object with mass m at a height h above a reference level, is:
[tex]U=mgh[/tex]Where g is the acceleration of gravity.
In the beginning, the car has a height of 10m and a speed of 10m/s. In the end, the car reaches the gas station at a height 15m and a unknown speed v.
Then, the initial speed and height are known, as well as the final height. Use the equation for the conservation of mechanical energy to isolate v_2, the unknown final speed of the car:
[tex]\begin{gathered} K_2+U_2=K_1+U_1 \\ \Rightarrow\frac{1}{2}mv_2^2+mgh_2=\frac{1}{2}mv_1^2+mgh_1 \\ \Rightarrow\frac{1}{2}v_2^2+gh_2=\frac{1}{2}v_1^2+gh_1 \\ \Rightarrow\frac{1}{2}v_2^2=\frac{1}{2}v_1^2+gh_1-gh_2 \\ \Rightarrow\frac{1}{2}v_2^2=\frac{1}{2}v_1^2+g\mleft(h_1-h_2\mright) \\ \Rightarrow v_2^2=v_1^2+2g\mleft(h_1-h_2\mright) \\ \\ \therefore v_2=\sqrt{v_1^2+2g(h_1-h_2)} \end{gathered}[/tex]Replace v_1=10m/s, h_1=10m, h_2=15m and g=9.8m/s^2 to find the speed of the car as it reaches the gas station:
[tex]\begin{gathered} v_2=\sqrt{(10\frac{m}{s})^2+2(9.8\frac{m}{s^2})(10m-15m)} \\ =1.4142...\frac{m}{s} \\ \approx1.4\frac{m}{s} \end{gathered}[/tex]Therefore, the car's speed as it coasts into the gas station on the other side of the valley will be approximately 1.4 meters per second.
B. MULTIPLE CHOICE. Choose the letter of the best answer2. the equation to calculate the momentum isa. p = mgb. p = mvc. p = mghd. p = mt
Answer:
b. p = mv
Explanation:
The momentum is the mass in motion, so it is calculated as the mass times the velocity. It means that the equation to calculate the momentum is
p = mv
Where m is the mass and v is the velocity of the object.
So, the answer is
b. p = mv
Sound travels faster on a cold day than on a warm day. Is this true or false?
ANSWER
False
EXPLANATION
Sound waves travel by causing vibrations in particles in the direction of the wave and these vibrations can travel through any state of matter (solid, liquid, gas).
This means that if the vibrations occur faster, the sound waves can also be transmitted faster. On a warm day, the temperature is higher, which means that molecules of air vibrate faster, resulting in faster transmission of sound waves.
Therefore, it is false that sound travels faster on a cold day than on a warm day.
The position of a particle is F(t) = 4.01²î - 3.0ĵ +2.03 km. (a) What is the velocity of the particle at 0 s and at 1.0 s? (b) What is the average velocity between 0 s and 1.0
The velocity of the particle at 0 s and at 1.0 and the average velocity between 0 s and 1.0
a)[tex]\vec{V}(0)=0 m / s[/tex]
[tex]\vec{V}(1)=(8 \hat{\imath}+6 \hat{k}) m / s[/tex]
b)[tex](4 \hat{\imath}+2 \hat{k}) \mathrm{m} / \mathrm{s}[/tex]
This is further explained below.
What is the average velocity?Generally, the equation for the Position is mathematically given as
F(t) = 4.01²î - 3.0ĵ +2.03 km.
Therefore
[tex]\begin{aligned}&\vec{r}(t)=\left(4 t^2 \hat{l}-3 \hat{\jmath}+2 t^3 \hat{k}\right) m \\&\vec{V}(t)=\frac{d \vec{r}(t)}{d t}=\left(8 t \hat{\imath}+6 t^2 \hat{k}\right) m / s\end{aligned}[/tex]
For A
[tex]\begin{aligned}&\vec{V}(0)=0 m / s \\&\vec{V}(1)=(8 \hat{\imath}+6 \hat{k}) m / s\end{aligned}[/tex]
For B
[tex]\text { Average velocity } &=\frac{\text { Total displacement }}{\text { time interval }} \\[/tex]
[tex]=\frac{\vec{r}(1)-\vec{r}(0)}{1} \\&=4 \hat{\imath}-3 \hat{\jmath}+2 \hat{k}-\left(-3 \hat{\jmath}^{\prime}\right) \\&=(4 \hat{\imath}+2 \hat{k}) \mathrm{m} / \mathrm{s}\end{aligned}[/tex]
In conclusion, The speed of the particle at 0 seconds and at 1.0 seconds, as well as the average speed between 0 seconds and 1.0 seconds
a)[tex]\vec{V}(0)=0 m / s[/tex]
[tex]\vec{V}(1)=(8 \hat{\imath}+6 \hat{k}) m / s[/tex]
b)[tex](4 \hat{\imath}+2 \hat{k}) \mathrm{m} / \mathrm{s}[/tex]
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What is the speed, in m/s, of a wave on a cord if it has a wavelength of 3.5 m and a period of 0.5 s?
The wavelength, period and velocity are related by the equation:
[tex]v=\frac{\lambda}{T}[/tex]where λ is the wavelength and T is the period. In this case the wavelength is 3.5 m and the period is 0.5 s; plugging these values we have:
[tex]\begin{gathered} v=\frac{3.5}{0.5} \\ v=7 \end{gathered}[/tex]Therefore, the speed of the wave is 7 m/s
Jake did an experiment in his science class. He found two of the same kind of beaker, and he filled them with different liquids. Then, he measured the volume and mass of each. The image below shows the two beakers.Based on the image, which of the following is the best conclusion? A. The liquids have the same volume and different masses. B. The liquids have different volumes and the same mass. C. The liquids have different volumes and different masses. D. The liquids have the same volume and the same mass.
Yellow liquid:
Mass = 470 g
Volume = 450ml
Pink liquid
Mass= 570 g
Volume= 450ml
Volumes are equal (450ml) and masses are different (470g and 570g)
A. The liquids have the same volume and different masses.
Toaster uses a nichrome heating coil and operates at 120 V. When the toaster is turned on at 20°C, the current in the cold coil is 1.5 A. When the coil warms up, the current has a value of 1.3 A. If the thermal coefficient of resistivity for nichrome is 4.5x10-4 1/Co, what is the temperature of the coil?Group of answer choices68oC490oC160oC360oC260oC
Given that the operating voltage is V = 120 V.
The initial temperature of the toaster is T1 = 20 degrees Celsius
The initial current in the coil is I1 = 1.5 A
The final current in the coil is I2 = 1.3 A
The thermal coefficient of resistivity for nichrome is
[tex]\alpha=4.5\times10^{-4}^{}\text{ }^{\circ}C^{-1}[/tex]We have to find the final temperature of the coil, T2.
The initial resistance of the coil is
[tex]\begin{gathered} R1=\frac{V}{I1} \\ =\frac{120}{1.5} \\ =80\Omega \end{gathered}[/tex]The final resistance of the coil is
[tex]\begin{gathered} R2\text{ =}\frac{V}{I2} \\ =\frac{120}{1.3} \\ =92.307\Omega \end{gathered}[/tex]The formula to calculate the final temperature of the coil is
[tex]\begin{gathered} \alpha=\frac{(R2-R1)}{R1(T2-T1)} \\ T2-T1=\frac{(R2-R1)}{\alpha\times R1} \\ T2=\frac{(R2-R1)}{\alpha\times R1}+T1 \end{gathered}[/tex]Substituting the values, the final temperature will be
[tex]\begin{gathered} T2=\text{ }\frac{92.307-80}{4.5\times10^{-4}\times80}+20 \\ \approx360^{\circ}\text{ C} \end{gathered}[/tex]Thus, the final temperature is 360 degrees Celsius.
When considering the Law of Universal Gravitation, the graph of force v. distance is _____.1)linear2)parabolic3)circular4)none of the above
According to Law of Universal Gravitation, the graph between Force and distance is hyperbolic as the force depends sqaure of the distance. Therefore, the correct option is (4).
state the dimension of energy in physics
hint: Energy = force × distance
Force = Mass * Acceleration = kg* m/s^2= MLT^-2
Distance = metres= L
Energy = MLT^-2 * L =ML^2T^-2
Hi can you help me understand how to do this?
From the information given,
initial velocity of parachute = 198 ft/s
We want to convert 198 ft/s to m/s
Recall,
1 m = 3.3 ft
x m = 198 ft
By crossmultiplying, we have
3.3x = 198
x = 198/3.3
x = 60
Thus,
198 ft/s = 60 m/s
Rate = 60 m/s
To determine the distance that the parachute will fall in 10 seconds, we would apply one of Newton's equations of motion which is expressed as
s = ut + 1/2gt^2
where
s = distance covered
u = initial velocity
t = time
g = acceleration due to gravity and its value is - 9.8m/s^2
From the information given,
t = 10
u = 60
By substituting these values into the formula, we have
s = 60 x 10 - 1/2 x 9.8 x 10^2
s = 600 - 490
s = 110
The distance covered by the parachute in 10s is 110 m
n a slap shot, a hockey player accelerates the puck from a velocity of 5 m/s to 30 m/s in the same direction. If the puck moves over a distance of 10 m during this process, what was the acceleration?
Given data
*The given initial velocity of the puck is u = 5 m/s
*The given final velocity of the puck is v = 30 m/s
*The given distance is s = 10 m
The formula for the acceleration is given by the kinematic equation of motion as
[tex]\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{(30)^2-(5)^2}{2\times10} \\ =43.75m/s^2 \end{gathered}[/tex]Hence, the acceleration of the puck is a = 43.75 m/s^2
Sheena can row a boat at 2.90 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 2.00 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head upstream at an angle of 25.0° from the direction straight across the river.What is her speed with respect to the starting point on the bank?How long does it take her to cross the river?How far upstream or downstream from her starting point will she reach the opposite bank? If upstream, enter a positive value and if downstream, enter a negative value.In order to go straight across, what angle upstream should she have headed?
Given
v = 2.90 mi/h
x = 1.20 mi
vs = 2.0 mi/h
θ = 25°
Procedure
Part a)
Velocity of the boat with respect to water stream is given as
[tex]\begin{gathered} v_b=v-v_s \\ v_b=(2.90\cdot\sin 25-2.00)\hat{j}+2.90\cdot\cos 25\hat{i} \\ v_b=-0.77\hat{j}+2.62\hat{i} \end{gathered}[/tex]magnitude of the speed is given as
[tex]\begin{gathered} v_b=\sqrt[]{0.77^2+2.62^2} \\ v_b=2.73mi/h \end{gathered}[/tex]Part b)
Time to cross the river is given as:
[tex]\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{1.2mi}{2.90\cdot\cos25} \\ t=0.46h \end{gathered}[/tex]Part c)
Distance moved by boat in downstream is given as
[tex]\begin{gathered} x=v_yt \\ x=-0.77\cdot0.46 \\ x=-0.354mi \end{gathered}[/tex]Part d)
In order to go straight, we must net speed along the stream must be zero
so we will have
[tex]\begin{gathered} v\sin \theta=v_s \\ 2.90\sin \theta=2.00_{} \\ \sin \theta=\frac{2.00}{2.90} \\ \theta=43.60^{\circ} \end{gathered}[/tex]A car is negotiating a flat circular curve of radius 50 m with a speed of 20 m/swithout slipping. The maximum centripetal force (provided by static friction) is 1.2 x10^4N. What is the mass of the car?1) 0.50 x 10^3 kg2) 1.0 x 10^3 kg3) 1.5 x 10^3kg4) 2.0 x 10^3 kg
We are given a car that is experiencing a centripetal force.
The formula for the force is given by:
[tex]F_c=\frac{mv^2}{r}[/tex]Where "m" is the mass, "v" is the velocity and "r" is the radius. Now we solve for the mass, first by multiplying both sides by r:
[tex]rF_c=mv^2[/tex]Now we divide by the velocity squared:
[tex]\frac{rF_c}{v^2}=m[/tex]Now we replace the known values:
[tex]\frac{(50m)(1.2\times10^4N)}{(20\frac{m}{s})^2}=m[/tex]Solving the operations:
[tex]1500kg=m[/tex]Therefore, the mass is 1500 kg.
17. A material that makes energy transfer difficult is one that is a good...Select one:a. radiator.b. insulator.c. conductor.d. convector.
b. insulator.
The others describe a type of transfer of energy
