A short peice of metal that act as an electrical safety device by melting when the large current passes through it. The melting of the metal stops the flow of current and provide safety.
This metal peice is known as fuse.
Hence, option 4 is the correct answer.
Only need question 9 done. BIG POINTS, need asap
Answer:
Explanation:
Given:
V = - 1.5 m/s
t = 8 s
___________
X - ?
Position:
X =V·t = - 1.5·8 = - 12 m
An engineer is designing the runway of an airport. Of the planes that will use the airport, the lowest acceleration rate is 3 m/s^2 and the lowest take off speed is 85 m/s. Assuming these minimum parameters are for the same plane, what is the minimum allowed length of the runway
Given,
The lowest acceleration, a=3 m/s²
The lowest takeoff speed, v=85 m/s
The initial velocity of the plane is u=0 m/s, as the plane starts accelerating from the rest.
The minimum allowed length of the runway must be equal to the distance traveled by plane with minimum parameters before taking off. That is the distance traveled by the plane when it accelerates from 0 m/s to 85 m/s.
From the equation of the motion,
[tex]v^2-u^2=2as[/tex]Where s is the minimum allowed length of the runway.
On substituting the known values in the above equation,
[tex]undefined[/tex]Car A and Car B travelling on a straight level track points X and Y are 3 km apart on the track. car A passes point X at precisely 10 o’clock travelling to the right at 10 m/s car B passes point Y 2 minutes after car A passes point X travelling to the left at 15 m/s calculate the time and position where car and car b will pass each other?Velocity is constant
First, let's draw a diagram of what is happening
We can solve this by turning this into a system of equations
[tex]d_a=10t[/tex]This equation states where vehicle A is, as it travels 10 m/s per second.
[tex]d_b=3000-15t-120(10)[/tex]We subtract 120(10) from 3000 because that is the distance Car A moves in the two minutes that Car B hasn't moved.
This gives us the 2 equations
da = 10t
& db = 1800 -15t
If we want to find where the distances are the same, we can set both equations to each other
10t = 1800 -15t
1800 = 25t
t = 72 + 120 (2 minutes where vehicle A was already moving) = 192 seconds
We can also find where exactly they crossed each other by solving for distance with one of the two equations.
da = 10(192) = 1920 meters.
A sailboat travels 15 km [E] and then 11 km [W]. What is the boat's final position?A. 26 kmB. 26 km [E]C. 26 km [W]D. 4.0 km [E]
We will have the following:
First, we will assume that movement to the east is a positive value and movement to the west is negative, so:
[tex]d=15-11\Rightarrow d=4[/tex]So, the boat's final position is 4.0 km east.
11. If the car were allowed to travel for 5 additional seconds, what would the totaldisplacement of the car be then? (In case it is difficult to see... at the beginning,t=Os and x=Om at the end t=5s and x=1m)
Answer:
2 m
Explanation:
We can see that the car travels 1 m in 5 seconds, if you allowed 5 additional seconds, it will travel 1 m more. So, the total displacement will be:
1 m + 1 m = 2 m
Because it travels 1 m for every 5 seconds.
Therefore, the answer is
2 m
Earth A. Open system B. Closed system C. Isolated system
ANSWER:
A. Open system
STEP-BY-STEP EXPLANATION:
The Earth constitutes an open material system, since it exchanges matter and energy with outer space. The Earth's atmosphere continuously releases molecules to the outside and receives micrometeorites (matter).
#1 please. I’m crying and need this done by tmr. I need you to hurry I’m so tired.
Given that the diameter of the wheel is d = 35 m.
Also, given that it completes 1 rev in 2 min 20 sec.
We have to find angular velocity, tangential velocity and centripetal acceleration.
The radius of the wheel will be
[tex]\begin{gathered} r\text{ =}\frac{d}{2} \\ =\frac{35}{2} \\ =17.5\text{ m} \end{gathered}[/tex]The tangential velocity is
[tex]\begin{gathered} v=\frac{2\pi r}{t} \\ =\frac{2\times3.14\times17.5}{2\times60+20} \\ =0.785\text{ m/s} \end{gathered}[/tex]The angular velocity will be
[tex]\begin{gathered} \omega=\frac{v}{r} \\ =\frac{0.785}{17.5} \\ =0.044\text{ rad/s} \end{gathered}[/tex]The centripetal acceleration is
[tex]\begin{gathered} a_c=\frac{v^2}{r} \\ =\frac{(0.785)^2}{17.5} \\ =0.035m/s^2 \end{gathered}[/tex]An astronaut and equipment has a total mass of 182.8 kg. While initially at rest inspace, the astronaut fires a 144.4 newton rocket for 1.190 seconds. What is theaverage speed of the astronaut (in km/h)?
Given data:
* The total mass of an astronaut and equipment is,
[tex]m=182.8\text{ kg}[/tex]* The force is given as,
[tex]F=144.4\text{ N}[/tex]* The time taken is,
[tex]t=1.19\text{ s}[/tex]Solution:
According to the Newton's second law,
[tex]F=ma[/tex]where a is the acceleration of the body,
Substituting the known values,
[tex]\begin{gathered} 144.4=182.8\times a \\ a=\frac{144.4}{182.8} \\ a=0.7899ms^{-2} \end{gathered}[/tex]By the kinematic equations,
[tex]v-u=at[/tex]where u is the initial velocity and v is the final velocity,
Substituting the known values,
[tex]\begin{gathered} v-0=0.7899\times1.19 \\ v=0.94ms^{-1} \end{gathered}[/tex]The average speed of the astronaut is,
[tex]\begin{gathered} s_{average}=\frac{v+u}{2} \\ =\frac{0.94+0}{2} \\ =0.47ms^{-1} \end{gathered}[/tex]The average speed of the astronaut in Km/h is,
[tex]\begin{gathered} s_{average}=0.47\times\frac{10^{-3}^{}^{}}{\frac{1}{60\times60}} \\ s_{average}=0.47\times60\times60\times10^{-3} \\ s_{average}=1.692kmh^{-1} \end{gathered}[/tex]Thus, the average speed of the astronaut is 1.692 km/h.
Your classmate’s mass is 65 kg and the table weighs 490 N. Calculate the normal force on the table by the floor. Show your work
mass = 65kg
table weighs = 490 N
F = ma
Where:
F = force
a = acceleration due gravity (9.8 m/s^2)
F = 65 x 9.8 = 637 N
Add both weights
F = 637 + 490 = 1,127 N
N = sum of forces
N= Normal force = 1,127 N
A standard rifle has a mass of 3.27 kilograms. If it were used to fire an "elephant gun" round of mass 67.5 grams at +625 meters per second, what would be the recoil velocity of the rifle? Include units in your answer. The answer must be in 3 significant digits.
ANSWER:
-12.9 m/s
STEP-BY-STEP EXPLANATION:
We have the following information:
[tex]\begin{gathered} m_r=3.27\text{ kg} \\ m_b=67.5\text{ g = 0.0675 kg} \\ V_{\text{fusil}1}=V_{\text{bala}1}=0 \\ V_{\text{bala}1}=625\text{ m/s} \end{gathered}[/tex]We can calculate the recoil velocity that would be the velocity of rifle 2, just like this:
[tex]m_r\cdot V_{\text{fusil}1}+m_b\cdot V_{\text{bala}1}=m_r\cdot V_{\text{fusil}2}+m_b\cdot V_{\text{bala}2}[/tex]Replacing and solving:
[tex]\begin{gathered} 3.27\cdot0+0.0675\cdot0=3.27\cdot V_{\text{fusil}2}+0.0675\cdot625 \\ 0=3.27V_{\text{fusil}2}+42.1875 \\ V_{\text{fusil}2}=-\frac{42.1875}{3.27} \\ V_{\text{fusil}2}=-12.9 \end{gathered}[/tex]The recoil velocity of the rifle is -12.9 m/s
A 912 kg truck, moving forward at 59.6 km/h, collides head-on into a stationary 618 kg car. If they stick together right after the impact, what is the magnitude of the velocity in m/s
Given:
The mass of the truck is m1 = 912 kg
The initial velocity of the truck is
[tex]\begin{gathered} v_{1i}=\text{ 59.6 km/h} \\ =59.6\times\frac{1000\text{ m}}{3600\text{ s}} \\ =\text{ 16.556 m/s} \end{gathered}[/tex]The mass of the car is m2 = 618 kg
The initial velocity of the car is
[tex]v_{2i}=\text{ 0 m/s}[/tex]To find the velocity of the truck and car after the collision.
Explanation:
According to the conservation of momentum,
[tex]\begin{gathered} m1v_{1i}+m1v_{2i}=\text{ \lparen m1+m2\rparen v}_f \\ v_f=\frac{m1v_{1i}+m1v_{2i}}{m1+m2} \end{gathered}[/tex]On substituting the values, the magnitude of velocity after the collision will be
[tex]\begin{gathered} v_f=\frac{(912\times\text{ 16.556\rparen -\lparen618}\times0\text{\rparen }}{(912+618)} \\ =\text{ 9.86 m/s} \end{gathered}[/tex]Thus, the magnitude of velocity after the collison is 9.86 m/s
An object of charge +2.0 x10^-3 C is located at the Cartesian origin, (0m,0m). A second object, of charge -4.0x10^-3 C, is located at a position (50m, 0m). Find the full magnitude and direction of the Electric force present on a test charge of +1C, located at a position (0m, 50m).
Explanation:
We can represent the situation with the following figure
Therefore, we need to calculate the magnitude of F1 and F2.
The magnitude of electric force between two charges is equal to
[tex]F=k\frac{q_1q_2}{r^2}[/tex]Where k = 8.98 x 10^9, q1 and q2 are the charges and r is the distance between the charges. The distance between the object of charge -4 x 10^(-3) C and the object of charge +1C can be calculated using the Pythagorean theorem as
[tex]\begin{gathered} \text{ Pythagorean Theorem: c=}\sqrt{a^2+b^2} \\ \\ \text{ Replacing a = 50 and b = 50} \\ c=\sqrt{50^2+50^2} \\ c=\sqrt{2500+2500} \\ c=\sqrt{5000} \\ c=70.71\text{ m} \end{gathered}[/tex]Therefore, the distance between the charges is 70.71 m.
Then, replacing q1 = -4 x 10^(-3) C, q2 = +1C and r = 70.71 m, on the initial equation, we get:
[tex]\begin{gathered} F1=(8.98\times10^9)\frac{(-4\times10^{-3})(+1)}{70.71^2} \\ \\ F1=7184\text{ N} \end{gathered}[/tex]To calculate F2, we need to replace q1 = +2 x 10^(-3) C, q2 = +1C, and r = 50 m, so
[tex]\begin{gathered} F2=(8.98\times10^9)\frac{(2\times10^{-3})(1)}{50^2} \\ \\ F2=7184\text{ N} \end{gathered}[/tex]Now, we need to calculate the resultant force, so we need to identify the x and y coordinates of each force and add them
[tex]\begin{gathered} F1x=F1\cos(45)=7184\cos45=5079.86 \\ F1y=F1\sin(45)=7184\sin45=-5079.86 \\ F2x=0 \\ F2y=F2=7184 \\ \\ \text{ Resultant force} \\ Fx=F1x+F2x \\ Fx=5079.86+0=5079.86\text{ N} \\ \\ Fy=F1y+F2y \\ Fy=-5079.86+7184=2104.14\text{ N} \end{gathered}[/tex]Finally, we can calculate the magnitude and direction of the force as follows
[tex]\begin{gathered} \text{ magnitude } \\ F=\sqrt{(Fx)^2+(Fy)^2} \\ F=\sqrt{(5079.86)^2+(2104.14)^2} \\ F=5498.40\text{ N} \\ \\ \text{ Direction} \\ \theta=\tan^{-1}(\frac{Fy}{Fx}) \\ \\ \theta=\tan^{-1}(\frac{2104.14}{5079.86}) \\ \\ \theta=\tan^{-1}(0.41) \\ \theta=22.5° \end{gathered}[/tex]Therefore, the magnitude of the electric force is 5498.40 N and the direction is 22.5 degrees.
An electron in a hydrogen atom jumps from orbit n = 1 to n = 2. Which of these has occurred in this process? a)a photon of energy 2.55 eV is absorbedb)a photon of energy 10.2 eV is absorbedc)a photon of energy 1.89 eV is emitted
The difference in energy between orbits n = 1 and n = 2 is 10.2 eV, and when an electron moves from a lower orbit to a higher one, it absorbs energy, or in this case, a photon.
Answer: b) a photon of energy 10.2 eV is absorbed
What is biology and how is it helpful?
Biology is the branch of science that is a study of life or living beings. Word biology is derived from Greek words, 'bios', meaning lif
Two ice skaters stand face to face and push off and travel directly away from each other. The boy has a velocity of 5 m/s and he weighs 750N. The girl weighs 480 N. What is the girls velocity?
Given:
Weight 1 = 750N
Weight 2 = 480 N
V1 = velocity 1 = 5 m/s
First, calculate the masses of each one:
w = mg
g= 9.8 m/s
W1 = m1 g
750 = m1 (9.8)
m1= 750 /9.8 = 76.5 kg
m2 = 480/9.8 = 49 kg
Both momentums (p) add up to zero
p = mv
p1 + p2 = 0
m1v1 + m2v2 = 0
(76.5) (5 ) + (49)(v2) = 0
382.5 +49v2 = 0
382.5 = -49v2
382.5/-49 = v2
v2 = -7.8 m/s ( negative since velocity is in the opposite direction of the boy)
Suppose a scientist is attempting to create a model to explain the properties and behavior of mathematical wavesWhich of the following mathematical relationships would be useful?
In order to find the correct answer, let's analyze the correct formula that relates the speed, wavelength and period (or frequency) of a wavelength:
[tex]\begin{gathered} v=\lambda\cdot f\\ \\ f=\frac{1}{T}\\ \\ v=\frac{\lambda}{T} \end{gathered}[/tex]Since it's more common to identify a wave by its frequency, the first formula is more used:
The speed is equal to the waveength tmultiplied by the frequency.
Correct option: B.
If a 140 g baseball strikes your head at 30 m/s and stops in 0.0015 s, what is the magnitude of the ball's deceleration?
mass = 140 g
speed = 30 m/s
time = 0.0015 s
i need help can you help me i need to get this right but i cant figure it out Drag each label to the correct graph.Match each description of an object’s motion with the position-time graph that represents it.not movingmoving with constantspeedspeeding upslowing down
The above diagram represents speeding down as distance travelled is decreasing with time.
Here, the object is not moving as distance remains the same with changing time.
Here, the object is moving with constant speed i.e. distance travelled is proportional to time.
Here, the object is speeding up as the distance travelled is increasing much faster.
While an elevator of mass 2325 kg moves upward, the tension in the cable is 38.8 kN. Assume the elevator is supported by a single cable. Forces exerted by the guide rails and air resistance are negligible.the acceleration of the elevator in the upward direction = 6.88 m/s2If at some point in the motion the velocity of the elevator is 1.20 m/s upward, what is the upward velocity of the elevator 4.00 s later?
Given data:
* The acceleration of the elevator is,
[tex]a=6.88ms^{-2}[/tex]* The initial velocity of the elevator is u = 1.2 m/s.
* The time taken by the elevator is t = 4 seconds.
Solution:
By the kinematics equation, the final velocity of the elevator is,
[tex]v-u=at[/tex]where v is the final velocity,
[tex]\begin{gathered} v-1.2=6.88\times4 \\ v-1.2=27.52 \\ v=27.52+1.2 \\ v=28.72\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of the elevator in the upward direction is 28.72 m/s in the upward direction.
10g bullet is fired into a 2.5kg ballistic pendulum and becomes embedded in it. If the pendulum raises a vertical distance of 8cm, calculate the initial speed of the bullet.
Answer:
19.8m/s
Explanation:
Total mass = m = 2.5kg + 10g = 2.5 + 0.01 = 2.51 kg
Height gained = h = 8cm = 0.08m
Gain in potential energy by pendulum + bullet = mgh = 2.51 * 9.8 * 0.08 = 1.96784 J
By conservation of energy, loss in kinetic energy of bullet = gain in potential energy by pendulum + bullet = 1.96784 J
[tex]\frac{1}{2} mv^{2} = 1.96784[/tex]
m = 10 g = 0.01kg
v= [tex]\sqrt{\frac{2*1.96784}{0.01} }[/tex]
v = 19.8385 m/s = 19.8m/s approximately
What affect does a quadrupling of the mass have a upon the acceleration of the objectcan you identify a set of two trials that support the answer to this question
Take into account that the Newton's second law is given by the following formula:
F = m·a
where:
F: force
m: mass
a: acceleration
when you solve the formula above for acceleration a, you obtain:
a = F/m
if the mass is quadruplied, you have:
a' = F/(4m) = 1/4 · F/m
a' = 1/4 · a
that is, the acceleation is reduced to one quarter of the initial acceleration.
A 2.00 m brass rod with a 1.00 mm diameter behaves like a spring under stretching or compression. If the elastic modulus of brass is 9.10 × 1010 N/m2, then the spring constant of the rod is
Given data
*The given length of the rod is l = 2.00 m
*The given diameter of rod is d = 1.0 mm = 1.0 × 10^-3 m
*The elastic modulus of brass is E = 9.10 × 10^10 N/m^2
The formula for the spring constant of the rod is given as
[tex]\begin{gathered} k=\frac{Ea}{l} \\ =\frac{E\times(\frac{\pi}{4}d^2)}{l} \end{gathered}[/tex]*Here a is the cross-sectional area of the rod
Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{9.10\times10^{10}\times(\frac{3.14}{4}\times(1.0\times10^{-3})^2)}{2.00} \\ =3.57\times10^4\text{ N/m} \end{gathered}[/tex]Hence, the spring constant of the rod is k = 3.57 × 10^4 N/m
3 bulbs are in series and the same 3 bulbs are in parallel with the same battery. Which battery will run out faster?A. SeriesB. The sameC. Not enough infoD. Parallel
D. Parallel
Explanation:When bulbs are arranged in parallel, they draw more current, because the full voltage passes through each of the three bulbs, which means more current will be drawn by each of the bulbs.
For the bulbs arranged in series, the same current (which is one-third of the supplied current will pass through each of them)
Since the bulbs arranged in parallel draws more current from the battery, they will run out the battery faster.
The parallel connection will run out the battery faster
Can you send the ray diagram where the image comes behind the mirror(spherical) both concave and convex
To find
The ray diagram for the image when it is behind the mirror.
Explanation
For concave mirror
For convex mirror,
A soccer player kicks a soccer ball with a mass of 0.35kg. The player kicks with a force of 1,304N, and the kick lasts for 0.03seconds. What is the change in momentum of the soccer ball during the kick
Given data
*The given mass of the soccer ball is m = 0.35 kg
*The player kicks with a force is F = 1304 N
*The time is t = 0.03 s
The formula for the change in momentum of the soccer ball during the kick is given as
[tex]\Delta p=F\times t[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \Delta p=1304\times0.03 \\ =39.12\text{ kg.m/s} \end{gathered}[/tex]Hence, the change in momentum of the soccer ball during the kick is 39.12 kg.m/s
Question for practice class work at homethe question is number #14
Question 14.
Given:
• Frequency, f = 78.0 kHz.
,• Wavelength, λ = 0.333 m
,• Speed of sound in air = 344 m/s
Let's find the wave speed and how much faster is the speed than the speed of sound in air.
To find the wave speed, apply the formula:
[tex]\lambda=\frac{v}{f}[/tex]Where:
• v is the wave speed in m/s.
,• f is the frequency in Hz.
,• λ is the wavelength in meters (m).
Rewrite the formula for v and solve.
We have:
[tex]\begin{gathered} v=\lambda *f \\ \\ v=0.333*78\times10^3 \\ \\ v=25974\text{ m/s} \end{gathered}[/tex]Therefore, the speed of sound in solid is 25974 m/s.
Now, to find how much faster it is, we have:
[tex]\frac{speed\text{ of sound in solid}}{speed\text{ of sound in air}}=\frac{25974\text{ m/s}}{344\text{ m/s}}=75.5[/tex]Therefore, it is 75.5 times faster than the speed of sound in air.
ANSWER:
• Speed of sound in solid: , 25974 m/s.
• 75.5, times faster than the speed of sound in air
A person stands 1000 feet from the base of a statue. The angle ofelevation to the top of the statue is 23 degrees. How tall is the statue?
b = base
b = 1000f t
angle = 23 degress
Procedure
[tex]\begin{gathered} \tan \theta=\frac{\text{opposite}}{\text{adjacent}} \\ \tan 23=\frac{h}{1000} \\ h=1000\cdot\tan 23 \\ h=424.47\text{ ft} \end{gathered}[/tex]The height of the statue would be h = 424.47 ft
Bob pushed a crate up a 3.0m long ramp to a truck bed with a force of 800n. The truck bed is 1.5m off the ground and the crate has a mass of 120kg calculate the following work output work input efficiency of the inclined plane
The output work can be calculated as the change in mechanical energy of the crate.
Since the height of the crate increased by 1.5 meters, the change in mechanical energy occurs by an increase in potential energy:
[tex]\begin{gathered} E_p=m\cdot g\cdot h \\ E_p=120\cdot10\cdot1.5 \\ E_p=1800\text{ J} \end{gathered}[/tex]The input work is given by the work done by Bob to move the crate, using a force of 800 N in a distance of 3 meters:
[tex]\begin{gathered} W=F\cdot d \\ W=800\cdot3 \\ W=2400\text{ J} \end{gathered}[/tex]The efficiency is given by the output energy over the input energy:
[tex]e=\frac{\text{output}}{\text{ input}}=\frac{1800}{2400}=0.75[/tex]So the efficiency is 75%.
Fred the frog is hopping from lily pad to lily pad in search of a good fly forlunch. If the lily pads are spaced 2,4 m apart, and Fred jumps with a speedof 5.0 m/s, taking 0.60 s to go from lily pad to lily pad, at what angle mustFred make each of his jumps? (Use 10 for g)
Answer:
36.87°
Explanation:
The frog is making a projectile motion, so the time of flight for each jump is calculated as
[tex]t=\frac{2v_0\sin\theta}{g}[/tex]Where vo is the initial velocity, g is the gravity and θ is the angle. Solving for sinθ, we get
[tex]\begin{gathered} tg=2v_0\sin\theta \\ \sin\theta=\frac{tg}{2v_0} \end{gathered}[/tex]Replacing t = 0.60 s, v0 = 5.0 m/s, and g = 10 m/s², we get
[tex]\begin{gathered} \sin\theta=\frac{(0.60\text{ s\rparen}(10\text{ m/s}^2\text{\rparen}}{2(5.0\text{ m/s\rparen }} \\ \sin\theta=0.6 \\ \theta=\sin^{-1}(0.6) \\ \theta=36.87 \end{gathered}[/tex]Therefore, the angle is 36.87°
How long will it take for a car accelerating at 4 meters/sec/sec to cover a distance of 500 meters?
Given:
The acceleration of the car is,
[tex]a=4\text{ m/s}^2[/tex]The distance is,
[tex]s=500\text{ m}[/tex]To find:
The time to cover the distance
Explanation:
When an object starts from rest with acceleration, the distance is,
[tex]\begin{gathered} s=\frac{1}{2}at^2 \\ t=\sqrt{\frac{2s}{a}} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\sqrt{\frac{2\times500}{4}} \\ t=\sqrt{250} \\ t=15.8\text{ s} \end{gathered}[/tex]Hence, the required time is 15.8 s.