Answer:
a.
[tex]horizontal=39.9[/tex] m/s
[tex]vertical=30.1[/tex] m/s
b.
[tex]t=5.009[/tex]
c.
[tex]y=27.7[/tex]
Explanation:
Lets write down what we were given.
Angle = 37°
Initial Velocity = 50 m/s
Displacement in x direction = 200 m
Take note:
I am having some trouble with the theta symbol so let theta = [tex]N[/tex]
Lets do question C first.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
Rearranging terms, we have
[tex]y=(tan(N)*x)-[\frac{g}{2(v_{0}cos(N))^{2} } ]x^{2}[/tex]
Now lets substitute our numbers in for the variables. Then simplify.
[tex]y=(tan37*200)-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[0.0030761]200^{2}[/tex]
[tex]y=150.7108-(0.0030761*40000)[/tex]
[tex]y=150.7108-123.0444[/tex]
[tex]y=27.7[/tex]
Now lets do question B.
Lets steal this from the last question.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
We can substitute [tex]t[/tex] for [tex]\frac{x}{v_{0}cos(N) }[/tex]
[tex]y=(v_{0}sin(N))*(t)-\frac{1}{2}g(t) ^{2}[/tex]
We can rewrite the equation as
[tex](v_{0}sin(N)(t)-\frac{1}{2}*(g(t)^{2})=y[/tex]
Now lets substitute our numbers in for the variables.
[tex](50sin(37)(t)-\frac{1}{2}*(9.81(t)^{2})=27.7[/tex]
After some painful algebra and factoring we get
[tex]30.09075115t-4.905t^{2}=27.6664[/tex]
Subtract [tex]27.6664[/tex] from both sides.
[tex]30.09075115t-4.905t^{2}-27.6664=0[/tex]
Use the quadratic formula to find the solutions.
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]
After some more painful algebra we get
[tex]t=5.00854263, 1.12616708[/tex]
1.126 does not make any sense so.
[tex]t=5.009[/tex]
Finally lets do question A.
Lets draw a triangle. We have the velocity which is the hypotenuse and we have the angle. From there we can solve for the opposite and adjacent sides.
Let [tex]A=horizontal[/tex] and [tex]O=vertical[/tex]
[tex]cos(37)=\frac{A}{50}[/tex]
[tex]A=39.9[/tex]
[tex]sin37=\frac{O}{50}[/tex]
[tex]O=30.1[/tex]
Can all rocks be dated with radiometer methods? Explain
Answer: No.
Explanation:
Radiometer dating is used on igneous rocks.
Unlike the other two rock types, sedimentary annd metamorphisis, all igneous rocks possess one specific age/ time of origin.
Ultraviolet light has shorter wavelengths and higherfrequencies than visible light.TRUEFALSE
Ultraviolet light has shorter wavelengths and higher frequencies than visible light.
The answer is TRUE
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." Explain the error in the student's claim. Provide an example of each quantity to support your answer.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." The error here is that acceleration is said to be done when either speed of that object changes or direction of that object changes. Hence , acceleration is not a scaler quantity.
Scalar quantities are quantities that are described only by a magnitude. They do not have a direction of action.
A vector quantity is defined as the physical quantity that has both directions as well as magnitude.
Acceleration is said to be occurred in two cases :
when the object changes its speed
or when the object changes its direction
since , acceleration depends upon both direction and magnitude ,hence it is a vector quantity not a scaler quantity.
for example : a stone attached to a string moving in a circular motion at a constant speed will be considered in accelerated motion because it is constantly changing its direction.
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A car going at 80 mph comes to a complete stop in 6 seconds. Calculate the acceleration
The acceleration is defined as:
[tex]a=\frac{v_f-v_0}{t}[/tex]where vf is the final velocity, v0 is the initial velocity and t is the time. In this case we have:
The initial velocity is 80 mph
The final velocity is 0 mph (since the car stops)
The time it takes to slow down is 6 seconds.
Before we can do the calculation, we need to convert the velocity to appropriate units; let's write the initial velocity in ft/s units. To do this we need to remember that 1 mile is equal to 5280 ft and one hour is equal to 3600 s, then we have:
[tex]80\frac{mi}{h}\cdot\frac{5280\text{ ft}}{1\text{ mi}}\cdot\frac{1\text{ h}}{3600\text{ s}}=117.33\frac{ft}{s}[/tex]Hence the initial velocity is 177.33 ft/s.
Now that we have all the values we need, we plug them in the equation for the acceleration:
[tex]\begin{gathered} a=\frac{0-117.33}{6} \\ a=-19.56 \end{gathered}[/tex]Therefore, the acceleration is -19.56 feet per second per second. Note: The minus sign indicates that the car is slowing down.
What word am I looking for I have already tried losing
We will have the following:
neutral object becomes charged by losing electrons.
When a 10 V battery is connected to a resistor, 5 A of current flows through the resistor. What is the resistor's value?
Given data
*The value of battery voltage is V = 10 V
*The current flows through the resistor is I = 5 A
The formula for the resistor is given by the Ohm's law as
[tex]R=\frac{V}{I}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}[/tex]Two drops of mercury each has a charge on 2.42 nC and a voltage of 293.97 V. If the two drops are merged into one drop, what is the voltage on this drop?
The electric potential is given by:
[tex]\begin{gathered} V=\frac{Kq}{r} \\ \end{gathered}[/tex]Let's find r first:
[tex]\begin{gathered} r=\frac{Kq}{V}=\frac{8.988\times10^9\cdot2.42\times10^{-9}}{293.97} \\ r\approx0.074m \end{gathered}[/tex]Now we can find the radius of the new drop:
[tex]r_t=2(r)=2(0.074)=0.148[/tex]So:
[tex]\begin{gathered} V=\frac{K(2q)}{r_t}=\frac{8.988\times10^9\cdot2(2.42\times10^{-9})}{(0.148)} \\ V=293.93V \end{gathered}[/tex]A roller coaster car begins its roll from the top of the tracks at a speed of2 meters per second. When it reaches the bottom of the 200-meter drop four seconds later, its speed is 22 meters per second. What was the averagespeed of the roller coaster ride in meters per second over the 200-meter drop
The speed of the car from the top of the track is,
[tex]u=2ms^{-1}[/tex]The distance traveled by the car is,
[tex]d=200\text{ m}[/tex]The speed of the car after 4 seconds is,
[tex]v=22ms^{-1}[/tex]Thus, the average speed of the car is,
[tex]v_{av}=\frac{u+v}{2}[/tex][tex]\begin{gathered} v_{av}=\frac{2+22}{2} \\ v_{av}=12ms^{-1} \end{gathered}[/tex]Thus, the average speed of the roller coaster car is 12 meter per second.
7. What is the velocity of a 850kg car after starting at rest when 13,000J of work is done to it.
Answer:
5.53 m/s
Explanation:
The work is equal to the change in the kinetic energy, so
[tex]\begin{gathered} W=\Delta KE \\ W=\frac{1}{2}m(v^2_f-v^2_i)^{}^{} \end{gathered}[/tex]Since the car starts at rest, the initial velocity vi = 0 m/s, so we can solve for the final velocity vf as follows
[tex]\begin{gathered} W=\frac{1}{2}mv^2_f \\ 2W=mv^2_f \\ \frac{2W}{m}=v^2_f \\ v_f=\sqrt[]{\frac{2W}{m}} \end{gathered}[/tex]So, replacing the work W = 13,000J and the mass m = 850kg, we get:
[tex]\begin{gathered} v_f=\sqrt[]{\frac{2(13,000J)}{850\operatorname{kg}}} \\ v_f=5.53\text{ m/s} \end{gathered}[/tex]Therefore, the velocity is 5.53 m/s
Suppose a 345-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.1 m from the ground to a branch.How much work, in joules, did the bird do on the snake? How much work, in joules, did it do to raise its own center of mass to the branch?
The work done in each case can be calculated with the change in potential energy of the body:
[tex]Work=m\cdot g\cdot h[/tex]The work done by the bird on the snake will use only the mass of the snake (in kg):
[tex]\begin{gathered} Work=0.075\cdot9.8\cdot2.1\\ \\ Work=1.5435\text{ J} \end{gathered}[/tex]The work done by the bird to raise its own center of mass will use only the bird mass (in kg):
[tex]\begin{gathered} Work=0.345\cdot9.8\cdot2.1\\ \\ Work=7.1\text{ J} \end{gathered}[/tex]A.Calculate the combined force of vector F ?B.Calculate the direction of the combined force vector F ?
Answer:
A. 282.93 N
B. 1.94 degrees
Explanation:
The combined force is found by first adding the three forces given.
We add the three forces by adding their x and y components separately and then combining the results to produce the total force,
The x component of a force is
[tex]\begin{gathered} \cos \theta=\frac{f_x}{F} \\ \Rightarrow f_x=F\cos \theta \end{gathered}[/tex]Therefore, x components of the forces is
[tex]F_x=120\cos 65+100\cos 25+200\cos (-45)[/tex]The y-component of the forces is
[tex]F_y=120\sin 120+100\sin 25+200\sin (-45)[/tex]Now evaluating the above two components gives
[tex]F_x=282.77N[/tex][tex]F_y=9.597N[/tex]Let us draw on big vector whose components are the above vectors.
The angle of the combined vector with respect to the x-axis is
[tex]\tan \theta=\frac{9.59}{282.77}[/tex][tex]\theta=\tan ^{-1}(\frac{9.59}{282.77})[/tex][tex]\boxed{\theta=1.94^o}[/tex]which is our answer!
The magnitude of the combined vector is
[tex]F=\sqrt[]{F^2_x+F^2_y_{}}[/tex][tex]F=\sqrt[]{(9.59)^2_{}+(282.77)^2_{}}[/tex][tex]\boxed{F=282.93N}[/tex]which is our answer!
Hence, to summerise:
A. 282.93 N
B. 1.94 degrees
nd in Atlanta you decide to drive around the city. You turn a corner and are driving up a steep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to a stop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching from the sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes an angle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’s information book tells you that the mass of your car is 1600 kg. You weigh 140 lbs. Will you fight the ticket
skid = 50 ft
Speed limit = 25 mph
angle = 25°
friction coefficient = 0.80
mass = 1600 kg
weight = 140 lbs
An object is dropped from rest out of the window of a building, and the time to hit the ground is found to be 5 seconds. The same object is then dropped from rest out of a window twice as high above the ground as the original window. The time it takes the object to hit the ground is closest to:
ANSWER:
7 s
STEP-BY-STEP EXPLANATION:
Given:
u = 0m/s
t = 5 sec
g = 9.8 m/s^2
The first thing is to calculate the height of the building, using the following formula:
[tex]\begin{gathered} s=ut+\frac{1}{2}gt^2 \\ \text{ Replacing} \\ s=0\cdot5+\frac{1}{2}\cdot9.8\cdot5^2 \\ s=122.5\text{ m} \end{gathered}[/tex]Now, we apply the same formula, but we substitute the double value of the distance and solve for t, just like this:
[tex]\begin{gathered} 2\cdot122.5=\frac{1}{2}\cdot9.8\cdot\: t^2 \\ 9.8\cdot t^2=245\cdot2 \\ t^2=\frac{490}{9.8} \\ t=\sqrt[]{50} \\ t=7.07\text{ sec} \\ t\approx7\text{ sec} \end{gathered}[/tex]The time it takes for the object to fall is 7 seconds.
Describe the mathematical relationship between the distance (d) and the attractive force (F) between protons and electrons.
The attractive force and the distance are inversely proportional.
[tex]F\propto\frac{1}{r}[/tex]This relation means that the attractive force decreases as the distance increases, and the attractive force increases as the distance decrease.
what is the velocity of the boat from point x to point y
The distance between the points X and Y is,
[tex]\begin{gathered} d=9\text{ km} \\ =9000\text{ m} \end{gathered}[/tex]The time taken to move the distance is,
[tex]\begin{gathered} t=12\text{ min} \\ =12\times60\text{ s} \end{gathered}[/tex]The velocity is given by,
[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{9000\text{ m}}{12\times60\text{ s}} \\ =12.5\text{ m/s} \end{gathered}[/tex]Hence the velocity is 12.5 m/s.
Full working out…….2.A vibrating mass-spring system has a frequency of 0.56 Hz. How much energy ofthis vibration is carried away in a one-quantum change?
ANSWER
3.7128 x 10⁻³⁴ J
EXPLANATION
The energy carried in a one-quantum change is the product of Planck's constant, h, and the frequency of vibration, f,
[tex]E=hf[/tex]Planck's constant is 6.63 x 10⁻³⁴ J*s and, in this case, the frequency of vibration is 0.56 Hz. So, the energy carried away is,
[tex]E=0.56Hz\cdot6.63\cdot10^{-34}J\cdot s=3.7128\cdot10^{-34}J[/tex]Hence, the energy carried away in a one-quantum change is 3.7128 x 10⁻³⁴ J.
How does the work needed to stretch a spring 2 cm compare to the work needed to stretch it 1 cm.A.Same amount of workB.twice the workC.4 times the work D.8 times the work
The work required to stretch a string is given by the following equation:
[tex]W=\frac{1}{2}kx^2[/tex]Where:
[tex]\begin{gathered} k=\text{ string constant} \\ x=\text{ distance the string is stretched} \end{gathered}[/tex]If the string is stretched 2 cm then we substitute the value of "x = 2" in the formula, we get:
[tex]W_2=\frac{1}{2}k(2)^2[/tex]Solving the square and simplifying:
[tex]W_2=2k[/tex]Now, if the string is stretched 1 cm we get:
[tex]W_1=\frac{1}{2}k(1)^2[/tex]Solving the operations:
[tex]W_1=\frac{1}{2}k[/tex]Now, we determine the quotient between W2 and W1:
[tex]\frac{W_2}{W_1}=\frac{2k}{\frac{1}{2}k}[/tex]Simplifying we get:
[tex]\frac{W_2}{W_1}=4[/tex]Now, we multiply both sides by W2:
[tex]W_2=4W_1[/tex]Therefore, the work required to stretch the string 2 cm is 4 times the work to stretch it 1 cm.
A) the frictional force F newtonsB)The resultant normal reaction of the surface on the metal block
Given:
The mass of the block is.
[tex]m=10\text{ kg}[/tex]The tension on the rope is,
[tex]T=100\text{ N}[/tex]The angle with the horizontal is,
[tex]\theta=60^{\circ}[/tex]The block is moving with constant speed.
as the block is moving with constant speed, the net force on the block will be zero.
Part (A)
we can write in the horizontal direction the component of the tension will be equal to the frictional force and we write,
[tex]\begin{gathered} T\cos 60^{\circ}=F \\ F=100\cos 60^{\circ} \\ F=50\text{ N} \end{gathered}[/tex]Hence the frictional force is 50 N.
\\
Part(B)
The resultant normal reaction will be,
[tex]\begin{gathered} N=T\sin 60^{\circ}-mg \\ =100sin60^{\circ}-10\times9.8 \\ =-11.4\text{ N} \end{gathered}[/tex]hence the resultant normal reaction is -11.4 N.
A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?
Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:
[tex]\begin{gathered} d=vt \\ =(3.1\frac{m}{s})(314s) \\ =973.4m \end{gathered}[/tex]That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:
[tex]\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=\frac{d}{2\pi} \\ =\frac{973.4m}{2(3.14...)} \\ =154.9...m \end{gathered}[/tex]The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:
[tex]a_c=\frac{v^2}{r}[/tex]Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:
[tex]a_c=\frac{(3.1\frac{m}{s})^2}{(154.9m)}=0.062\frac{m}{s^2}[/tex]Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.
When you step off a bus moving at 2 m/s, your horizontal speed when you meet the ground isA) zero.B) less than 2 m/s but greater than zero.C) about 2 m/s.D) greater than 2 m/s.
ANSWER:
C) about 2 m/s.
STEP-BY-STEP EXPLANATION:
While step off the bus, it acquires a vertical component of velocity, but it still has the initial horizontal component of velocity due to the movement of the bus.
Which means that the velocity is either 2 m/s or about 2 m/s
A 20.0 kg penguin slides at a constant velocity of 3.3 m/s down an icy incline. The incline slopes above the horizontal at an angle of 6.0°. Determine the coefficient of kinetic friction.
The coefficient of kinetic friction down the slope is 0.105.
What is kinetic friction?Kinetic friction is the friction that exists or acts between the surfaces of one object moving over another.
The kinetic frictional force of an object moving on an inclined plane is give by the formula below:
Kf = μk * mg *cosθ
where;
μk = coefficient of kinetic friction.
mg cosθ = component of the weight perpendicular to the inclined plane
θ = angle of inclination
For an object moving at a constant velocity, the component of the weight down the slope (mg sinθ) is equal to the kinetic frictional force.
Hence, μk * mg *cosθ = mg sinθ
μk = mg sinθ / mg *cosθ
μk = tan θ
μk = tan 6.0
μk = 0.105
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Need help 82x2682 please
ANSWER:
STEP-BY-STEP EXPLANATION:
We have the following multiplication:
[tex]undefined[/tex]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.
a 298 kg boat is being propelled forward with a force of 2,365 N. What is the acceleration of the boat if it has a resistance force (rewarded) due to wind and water of 878 N? (Write answer as a 2 digit number)
The acceleration of the boat is 4.9m/s²
Mass of boat= 298 kg
Forward force= 2365 N
Resistance force= 878N
We need to apply the concept of laws of motion
Net force= Forward force- Resistance force
Net force= 2365-878 N
= 1487 N
Net force= mass x acceleration
2365= acceleration x 298
acceleration = 4.9 m/s²
Therefore, the acceleration of the boat is 4.9 m/s²
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Does scarcity effect everyone?
Answer: Scarcity affects society in every way. First and foremost, scarcity affects the way that individuals make choices. Time and money are two examples of scarce resources that we make choices with every day.
A ski jumper competing for an Olympic gold metal wants to jump
a horizontal distance of 149 meters. The takeoff point of the ski
jump is at a height of 38.0 meters. With what horizontal velocity
must he leave the jump in order to travel 149 meters?
19.25 m/s is horizontal velocity must he leave the jump in order to travel 149 meters .
How fast is horizontal moving?Standard definitions of horizontal velocity include miles per hour and meters per second, which are horizontal displacement times time. The distance an object has traveled since its origin is simply referred to as displacement.
How can one calculate vertical velocity using horizontal velocity?V * cos() equals the horizontal velocity component Vx. V * sin() is equal to the vertical component of velocity, Vy.
Time before landing = sqrt ( 2 x height / gravity ), sqrt ( 2 x 38/ 9.81) = 7.75
distance / time = avg speed
149/ 7.75 ≅ 19.25 m/s
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How long would it take to pass 700 C of charge through a toaster drawing 10 A of current? How many electrons would pass through the toaster in this time?
Given that the charge of the toaster is q =700 C
The current of the toaster is I = 10 A
We have to find the time and number of electrons.
Time can be calculated by the formula,
[tex]t=\frac{q}{I}[/tex]Substituting the values, the time will be
[tex]\begin{gathered} t=\frac{700}{10} \\ =70\text{ s} \end{gathered}[/tex]The number of electrons can be calculated by the formula,
[tex]n=\frac{q}{e}[/tex]Here, n is the number of electrons
and e is the charge of the electron whose value is
[tex]1.6\text{ }\times10^{-19}\text{ C}[/tex]Substituting the values, the number of electrons will be
[tex]\begin{gathered} n\text{ = }\frac{700}{1.6\times10^{-19}} \\ =4.375\text{ }\times10^{-17} \end{gathered}[/tex]Write a 5 page research on the topic 'How do migrating birds / animals find their direction? (Two birds / animals)'. Need this ASAP.
Answer: Birds migrate to move from areas of low or decreasing resources to areas of high or increasing resources. The two primary resources being sought are food and nesting locations. Here’s more about how migration evolved.
Birds that nest in the Northern Hemisphere tend to migrate northward in the spring to take advantage of burgeoning insect populations, budding plants and an abundance of nesting locations. As winter approaches and the availability of insects and other food drops, the birds move south again. Escaping the cold is a motivating factor but many species, including hummingbirds, can withstand freezing temperatures as long as an adequate supply of food is available.
Explanation: sorry about the anther
A person pushes a 500 kg crate with a force of 1200 N and the crate accelerates at .5 m/s^2. What is the force of friction acting on the crate?
The force of friction acting on the crate is 950 N.
What is force of friction?Force of friction is defined as the force that opposes the motion of an object when two surfaces are in contact.
The frictional force on the object is determined by applying Newton's second law of motion as shown below.
F - Ff = ma
where;
F is the applied force = 1200 NFf is the frictional forcem is the mass of the crate = 500 kga is the acceleration of the crate = 0.5 m/s²1200 - Ff = 500(0.5)
1200 - Ff = 250
Ff = 1200 - 250
Ff = 950 N
Thus, the force of friction acting on the crate preventing the motion of the crate is determined by applying Newton's second law of motion.
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If the planet Mercury has a mass of planet 3.3×10²³ kg and a radius of 2400 km - calculate the magnitude of the gravitational field on its surface?
The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
Explanation:The mass of mercury, m = 3.3×10²³ kg
The radius, r = 2400 km
r = 2400 x 1000m
r = 2.4 x 10⁶m
Note that the magnitude of the gravitational field on the surface of the planet is the acceleration due to gravity on that planet
It is given by the formula:
[tex]g=\frac{Gm}{r^2}[/tex]Substitute the values of G, m, and r into the formula above
[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{2.2\times10^{13}}{5.76\times10^{12}} \\ g=3.82m/s^2 \end{gathered}[/tex]The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²