The Time that elapses before the car is traveling at 3.7 m/s is 4.8 s.
What is time?Time is a measure of non-stop, consistent change in our surroundings, usually from a specific viewpoint.
To calculate the time require for the to be traveling at 3.7 m/s, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = Timev = Final velocityu = Initial velocitya = AccelerationFrom the question,
Given:
v = 3.7 m/su = 19.6 m/sa = - 3.33 m/s²Substitute these values into equation 1
t = (19.6-3.7)/-3.33t = -15.9/-3.33t = 4.8 sHence, the time need for the car to start traveling at 3.7 m/s is 4.8 s.
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Your friend is flying a hot air balloon and you are recording them from below. When you start filming, the hot air balloon is at an angle of 63 degrees above you while staying 100 feet away from where it launched. After 5 seconds, the angle has increased to 67.9 degrees. How fast is your friend rising in feet per seconds and miles per hour.
Answer:
10 ft/s and 6.818 mi/hr
Explanation:
We can represent the situation with the following diagram:
When the angle is 63 degrees, the value of h can be calculated using a trigonometric function as:
[tex]\tan 63=\frac{h_i}{100}[/tex]Because h is the opposite side and 100 ft is the adjacent side. Solving for h, we get:
[tex]\begin{gathered} h_i=100\times\tan 63 \\ h_i=196.26\text{ ft} \end{gathered}[/tex]In the same way, when the angle is 67.9 degrees, we can calculate the height as follows:
[tex]\begin{gathered} \tan 67.9=\frac{h_f}{100} \\ h_f=100\times\tan 67.9 \\ h_f=246.27ft \end{gathered}[/tex]Now, we can calculate the speed in ft per second as follows:
[tex]s=\frac{h_f-h_i}{t}=\frac{246.27ft-196.24ft}{5s}=10\text{ ft/s}[/tex]Finally, 1 mile = 5280 ft and 1 hour = 3600 seconds, so we can convert to miles per hour as:
[tex]10\text{ ft/s }\times\frac{1\text{ mile}}{5280\text{ ft}}\times\frac{3600}{1\text{ hour}}=6.818\text{ mi/hr}[/tex]Therefore, the answers are:
10 ft/s and 6.818 mi/hr
9. The graph is a plot of the velocity versus time for an objectmoving in a straight line. The x position of the objectat t = 0 seconds is 0 meters. At what time after t = 0 secondsdoes the object again pass through its initial position?(a) 1 second(b) Between 1 and 2 seconds(c) 2 seconds(d) 3 seconds
Velocity:
The velocity in simple word, is defined as the change in the position of the particle with respect to the time. The velocity is considered as a vector quantity as it has both the magnitude and direction in it. It can be measured or calculated in meter per second.
At t = 2 seconds, the speed of an object is -10 m/s. As it passes through its initial position. Hence the correct answer is (2)
SORRY IF ITS IN THE WRONG SUBJECT. I NEED SOMEONE ANSWER ASAP. text to answer the question. Most games that are played with a standard deck of playing cards are called trick games. In such a game, each player will take a turn playing a card, and whoever plays the winning card takes them all. These cards make up a trick, which the winner puts face-side down in a stack before playing the first card for the next round. Would the information in the text be an effective source to help answer the research question "What are the most popular magic tricks? (1 point) O No, because magic tricks that use playing cards are unpopular. O No, because the source is about card games instead of magic tricks. OYes, because the source is written by an expert on the subject. OYes, because the source explains a type of trick.
Answer: the answer should be "No, because the source is about card games instead of magic tricks."
Explanation: sorry for answering 2 weeks later
Answer:
B
Explanation:
Hi I do know the answer to this it’s be answered for me twice already just looking for a bit of a deeper explanation physical and why and how?
4.1
The free body diagram is shown below
Newton's first law of motion states that a body at rest or in motion would continue to be at rest or in motion unles an external force acts on it causing it to start or stop moving.
Sum of forces in the horizontal direction, Fx = - Fr + F
Sum of forces in the vertical direction, Fy = N + FSinθ - mg
Since the motion is along the horizontal axis, There would not be an acceleration in the y axis. Thus, Fy = 0
Force = mgCosθ
4.3) From the information given,
mass, m = 50kg
Force =
Forcce = 50 x 9.81Cos20
4.4)
Fy = 0
N + FSinθ - mg = 0
N = mg - FSinθ
N = 50 * 9.81 - FSin20
N = 490.5 - FSin20
N = 490.5 - 0.342F
4.5)
Frictional force = coefficient of friction x normal force
From the information given,
coefficient of friction = 0.4
Frictional force = 0.4(490.5 - 0.342F)
Which choice is a valid way to construct a motion diagram?1) Add together the average speeds of the various objects in motion.2) Take a series of photographs at equal time intervals of a moving object, perpendicular to the direction of motion; overlay the images to see how the position changes with time.3) Draw vectors to represent the speeds involved.4) Add vectors in a head-to-tail manner to determine the resultant vector.
A motion diagram displays the location of an object at various equally spaced time intervals in the same diagram. This allows us to visualize the motion of the object.
From the described procedures, the one which would be useful to make a motion diagram, is:
Option 2) Take a series of photographs at equal time intervals of a moving object, perpendicular to the direction of motion; overlay the images to see how the position changes with time.
Please helpppppppppp - the diagram shows the trajectory of a ball that is thrown horizontally from a top of a building. The ball’s vertical and horizontal velocity vectors along with the resultant factors are also indicated if the ball takes three seconds to reach the ground how fast is it moving by the time it reached the ground
A uniform 500 N/C electric field points in the positive y-direction and acts on an electron initially at rest. After the electron has moved4.00 cm in the field, what is the energy of electron in eV?
Given data:
* The electric field in the y-direction is,
[tex]E=500\text{ N/C}[/tex]* The distance traveled by the electron is,
[tex]\begin{gathered} d=4\text{ cm} \\ d=0.04\text{ m} \end{gathered}[/tex]Solution:
The work done in terms of electric field is,
[tex]W=\text{Edq}[/tex]where q is the charge on an electron,
Substituting the known values,
[tex]\begin{gathered} W=500\times0.04\times1.6\times10^{-19}\text{ J} \\ W=\frac{500\times0.04\times1.6\times10^{-19}}{1.6\times10^{-19}}\text{ eV} \\ W=500\times0.04\text{ eV} \\ W=20\text{ eV} \end{gathered}[/tex]This work done is stored in the charge in form of energy.
Thus, the energy of the electron in eV is 20 eV.
Two objects a distance apart are experiencing 40 N of force. How much force wouldthere be if you DOUBLED the distance between them?
We know that the masses experience a 40 N force between them, which is a gravitational force due to their mass. So, if we double their distance between them, the force will decrease due to Newton's Gravitational Law.
[tex]F=G\cdot\frac{m_1\cdot m_2}{d^2_{12}}[/tex]Let's use 2d.
[tex]F=G\cdot\frac{m_1\cdot m_2}{(2d)^2}=G\cdot\frac{m_1\cdot m_2}{4d^2}[/tex]As you can notice, the force would be divide by 4, so let's do that.
[tex]F=\frac{40N}{4}=10N[/tex]Therefore, if we double their distance, their force would be 10 N.A falcon with a mass [tex]m_{1}= 1.21kg[/tex] is diving at a speed of [tex]v_{1}=25.8m/s[/tex] at an angle of[tex]tetha=39.7*[/tex] below horizontal. A pigeon whose mass is [tex]m_{2}=0.62kg[/tex] is flying the the positive x direction at a speed of [tex]v_{2}=8.6m/s[/tex]. The falcon catches the pigeon, and they move as one. Neglect gravity and air resistance.
(a) write an expression for the x component of the final velocity
(b) write an expression for the y component of the final velocity
(c) what is the magnitude, in meters per seconds, of the final velocity?
(e) what is the angel, in degrees below the horizontal, that the final velocity makes with the x axis?
a ) The expression for the x component of the final velocity = 16 m / s
b ) The expression for the y component of the final velocity = 10.92 m / s
c ) The magnitude of final velocity = 19.37 m / s
m1 = 1.21 kg
m2 = 0.62 kg
Along x-axis,
u1 = u cos θ
u1 = 25.8 cos 39.7°
u1 = 19.87 m / s
u2 = 8.6 m / s
According to law of conservation of momentum
m1u1 + m2u2 = ( m1 + m2 ) vx
( 1.21 * 19.87 ) + ( 0.62 * 8.6 ) = ( 1.21 + 0.62 ) vx
vx = ( 24 + 5.3 ) / 1.83
vx = 16 m / s
Along y-axis,
u1 = u sin θ
u1 = 25.8 sin 39.7°
u1 = 16.5 m / s
u2 = 0
m1u1 = ( m1 + m2 ) vy
( 1.21 * 16.5 ) = ( 1.21 + 0.62 ) vy
vy = 19.98 / 1.83
vy = 10.92 m / s
v = √ vx² + vy²
v = √ 16² + 10.92²
v = √ 256 + 119.25
v = √ 375.25
v = 19.37 m / s
tan θ = vy / vx
tan θ = 10.92 / 16
tan θ = 0.68
θ = 34.22°
Therefore,
a ) The expression for the x component of the final velocity = 16 m / s
b ) The expression for the y component of the final velocity = 10.92 m / s
c ) The magnitude of final velocity = 19.37 m / s
d ) The angle below the horizontal, that the final velocity makes with the x axis = 34.22°
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The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?Hz(b) What is its wavelength in glass that has an index of refraction of 1.63?nm(c) What is its speed in the glass?Mm/sNeed Help?Read ItWatch ItSubmit Answer
ANSWER:
(a) 4.74*10^14 Hz
(b) 388.22 nm
(c) 184 Mm/s
STEP-BY-STEP EXPLANATION:
We have the following information:
[tex]\begin{gathered} \lambda=632.8\text{ nm} \\ n=1.63 \end{gathered}[/tex](a)
To calculate the frequency we use the following formula:
[tex]\begin{gathered} f=\frac{c}{\lambda} \\ c=3\cdot10^8\text{ m/s} \\ \lambda=632.8\text{ nm }=632.8\cdot10^{-9}\text{ m} \\ \text{replacing:} \\ f=\frac{3\cdot10^8}{632.8\cdot10^{-9}} \\ f=4.74\cdot10^{14}\text{ Hz} \end{gathered}[/tex](b)
In this case, we apply the following:
[tex]\begin{gathered} \lambda_1=\frac{\lambda}{n} \\ \text{ replacing} \\ \lambda_1=\frac{632.8}{1.63} \\ \lambda_1=388.22\text{ nm} \end{gathered}[/tex](c)
To calculate the speed it would be:
[tex]\begin{gathered} v=\frac{c}{n} \\ \text{ replacing} \\ v=\frac{3\cdot10^8}{1.63} \\ v=1.84\cdot10^8 \\ v=184\text{ Mm/s} \end{gathered}[/tex]A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum
Given data
*The given mass of the pendulum is m = 3 kg
*The given height is h = 0.3 m
The formula for the maximum speed of the pendulum is given as
[tex]v_{\max }=\sqrt[]{2gh}[/tex]*Here g is the acceleration due to the gravity
Substitute the values in the above expression as
[tex]\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}[/tex]Hence, the maximum speed of the pendulum is 2.42 m/s
what is the greatest mass of groceries that can be lifted safely with this bag given that the bag is raised with an acceleration of 1.80 m/s^2
We will have the following:
According to the image:
First, we remember that:
[tex]F=m\cdot a[/tex]Now, we will determine he maximum mass will be:
[tex]52.0N=m\cdot(1.8m/s^2)\Rightarrow m=\frac{^{}52.0N}{1.80m/s^2}[/tex][tex]\Rightarrow m=\frac{260}{9}kg\Rightarrow m\approx28.9kg[/tex]So, the maximum mass will be approximately 28.9 kg.
A 1000-watt kettle is connected to a 220-volt power source. Calculate the resistance of the kettle
The power of kettle is given as,
[tex]P=\frac{V^2}{R}[/tex]Plug in the known values,
[tex]\begin{gathered} 1000\text{ W=}\frac{(220V)^2}{R} \\ R=\frac{(220V)^2}{1000\text{ W}}(\frac{1\text{ ohm}}{1V^2W^{-1}})_{} \\ =48.4\text{ ohm} \end{gathered}[/tex]Thus, the resistance of the kettle is 48.4 ohm.
What is the pH of a pure WATER solution if the concentration of H+ in water is 10-7 ?
Given data:
Comcentration of H+ ion in water,
[tex]\lbrack H^+\rbrack=10^{-7}\text{ }\frac{mol}{l}[/tex]The pH value is given as,
[tex]pH=-\log _{10}\lbrack H^+\rbrack_{}[/tex]Substituting all known values,
[tex]pH=-\log _{10}\lbrack10^{-7}\rbrack[/tex]Since,
[tex]\log _{10}(a^b)=b\times\log _{10}(a)[/tex]Therefore,
[tex]\begin{gathered} pH=-7\times(-\log _{10}\lbrack10\rbrack) \\ =7\log _{10}\lbrack10\rbrack \end{gathered}[/tex]Since,
[tex]\log _{10}\lbrack10\rbrack=1[/tex]Therefore,
[tex]\begin{gathered} pH=7\times1 \\ =7 \end{gathered}[/tex]Therefore, pH of a pure water solution is 7.
Imagine an asteroid at rest in space that cracks into two pieces. The first piece moves at 1.4 times the speed of the second piece. Calculate the ratio of the first piece’s mass to the second piece’s mass
Since the total linear momentum of the system was 0 before the asteroid cracking into two pieces, then the magntude of the linear momentum of each piece must be the same after the cracking.
Then:
[tex]m_1v_1=m_2v_2[/tex]If the speed of the first piece is 1.4 times the speed of the second, then:
[tex]\begin{gathered} v_1=1.4v_2 \\ \Rightarrow m_1\times1.4v_2=m_2v_2 \\ \Rightarrow1.4m_1=m_2 \\ \Rightarrow m_1=\frac{m_2}{1.4} \\ \Rightarrow\frac{m_1}{m_2}=\frac{1}{1.4} \end{gathered}[/tex]Then, the ratio of the first piece's mass to the second piece's mass is 1:1.4
Please help me!A boy on a skateboard travels the first 600m of a trip at an average speed of 2 m/s. He then travels the next 800m in 200s and spends the last 100s at a speed of 5 m/s. Find the average speed of the bicyclist for this trip.
average speed = total distance / total time
speed x time = distance
part 1
D1= 600 m, s1=2m/s
Part 2
d2=800m, t2=200s
Part 3
s3=5m/s ,t3=100s
time1 = Distance1/speed1 = 600m/2m/s = 300s
Distance3 = speed3 x time 3 = 5 m/s x 100s = 500m
Total distance = d1+d2+d3 = 600 + 800 + 500 = 1900m
Total time = t1+t2+t3 =300 + 200 + 100 = 600 s
Avg speed = 1900m/600s = 3.17 m/s
The vertices of two rectangles are A(−5,−1),B(−1,−1),C(−1,−4),D(−5,−4) and W(1,6),X(7,6),Y(7,−2),Z(1,−2). Compare the perimeters and the areas of the rectangles. Are the rectangles similar? Explain.Perimeter of ABCD: , Area of ABCD: Perimeter of WXYZ: , Area of WXYZ:
In order to compare the perimeters and areas, let's first find two adjacent sides of each rectangle.
From ABCD, let's calculate AB and BC:
A and B have the same y-coordinate, so the length is the difference in x-coordinate:
AB = -1 - (-5) = -1 + 5 = 4
B and C have the same x-coordinate, so the length is the difference in y-coordinate:
AB = -1 - (-4) = -1 + 4 = 3
Therefore the perimeter and area are:
[tex]\begin{gathered} P=4+3+4+3=14 \\ A=4\cdot3=12 \end{gathered}[/tex]Now, for rectangle WXYZ, let's use WX and XY:
W and X have the same y-coordinate, so the length is the difference in x-coordinate:
WX = 7 - 1 = 6
X and Y have the same x-coordinate, so the length is the difference in y-coordinate:
XY = 6 - (-2) = 6 + 2 = 8
So the perimeter and area are:
[tex]\begin{gathered} P=6+8+6+8=28 \\ A=6\cdot8=48 \end{gathered}[/tex]In order to check if the rectangles are similar, let's check the following relation:
[tex](\frac{P_1}{P_2})^2=\frac{A_1}{A_2}[/tex]So we have:
[tex]\begin{gathered} (\frac{14}{28})^2=\frac{12}{48} \\ (\frac{1}{2})^2=\frac{1}{4} \\ \frac{1}{4}=\frac{1}{4}\text{ (true)} \end{gathered}[/tex]Since the relation is true, so the rectangles are similar.
It takes 10 J of energy to move a 2 C charge from point A to point B. What is the potential difference between points A and B?Group of answer choices20 V0.2 V5 VNone of the above
Take into account that potential difference is defined as the quotient between the energy requierd to move a charge a from a point A to a point B. Use the following formula:
[tex]\Delta V=\frac{\Delta U}{q}[/tex]where ΔV is the potential difference, ΔU is the energy and q is the charge.
In this case, ΔU = 10J and q = 2C. By replacing these values into the previous formula and by simplifying you obtain:
[tex]\Delta V=\frac{10J}{2C}=5V[/tex]where 1J/C = 1V (for the units of potential difference)
Hence, the potential difference is 5V
The postal service will not ship goods over 50 lbs without a special label. Donovan wants to estimate the weight of his package so he doesn't exceed the weight limit. He has a cast iron skillet that weighs 5.67 lbs, a dictionary that weighs 8.34 lbs, and a set of dishes that weighs 37.88 lbs. What will the estimated weight of Donovan's package be if he rounds each item to the nearest pound before totaling the weight?
52 Lb
Explanation
Step 1
round each itme to the nearest pound.
in this case, Rounding a price to the nearest pound is the same as rounding a decimal to the closest whole number
so
for example, If the price is $3.80 you can round up to $4 because the number in the tenths position is 8. The closest whole number to 3.8 is 4
hence
[tex]\begin{gathered} cast\text{ iron=5.67 Lbs}\rightarrow6\text{ LB} \\ dictionary=8.34\text{ Lb}\rightarrow8\text{ Lb} \\ a\text{ set of disehes=}37.88\text{ Lb}\rightarrow38\text{ Lb} \end{gathered}[/tex]Step 2
now, add the cost of the items to find the weight of the package
[tex]\begin{gathered} Weight_{package}=Weight_{cast}+Weight_{dictironay}+Weight_{dishes}\text{ } \\ \text{replace} \\ Weight_{package}=6\text{ lb+8 lb +38 lb=52 lb} \end{gathered}[/tex]therefore, the answer is
52 Lb
I hope this helps you
A 100-N ball suspended by a rope A is pulled to one side horizontally by another rope B and supported so that the rope A makes an angle of 30º with the vertical wall (see figure). Find the tensions of the ropes A and B. You solve it for me by fi uwu
Tension in A rope = 58.115 N
Tension in B rope = 116.23 N
What is tension?Tension is defined as the force transmitted through a rope, cord, or wire when pulled by forces acting from opposite sides. Tension is transmitted along the length of the wire, drawing energy evenly into the bodies at both ends. T = mg + ma
Where;
T = tension, (N)
m = mass (kg)
g = gravitational force, 9.8 m/s²
A = acceleration (m/s²)
As, F = mg
100 = m x 9.8
m = [tex]\frac{100}{9.8}[/tex]
m = 10.20 kg
Let the tension in A rope be T₁ and in the B rope be T₂ which is making angle of 30⁰
The vertical component of tension T₂ will balance the weight
= T₂ cos 30 = 10.20 x 9.8
T₂ = [tex]\frac{99.96}{cos 30}[/tex]
Since cos 30 = [tex]\frac{\sqrt{3} }{2}[/tex]
T₂ = [tex]\frac{99.96}{0.86}[/tex]
T₂ = 116.23 N
The horizontal component of T₂ will balance T₁
T₂ sin 30 = T₁
116.23 sin 30 = T₁
T₁ = 116.23 × [tex]\frac{1}{2}[/tex]
T₁ = 58.115 N
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A car stereo pulls 2.89 A from a car's battery. If the battery has a voltage of 12 V, how much power does it use?
In order to calculate the power, we can multiply the voltage and the current:
[tex]P=I\cdot U[/tex]So we have:
[tex]\begin{gathered} P=2.89\cdot12\\ \\ P=34.68\text{ W} \end{gathered}[/tex]Therefore the power is 34.68 W.
. A person pushes on a hockey puck with their stick at an angle so the vertical force is 22 N[down] and the horizontal force is 45 N [forward]. Assume the ice is frictionless.a) What is the actual force the hockey player transmits to the puck?b) What is the work done by the person pushing the hockey stick if they push the puck for 3.0s as it moves with a constant velocity of 22 m/s [forward]?c) What is the significance of the fact that both the horizontal force and motion are bothforwards?
50.08 Newton is the actual force the hockey player transmits to the puck and 3305.93 Joule is the work done by the person pushing the hockey stick if they push for 3.0s as it moves with a constant velocity of 22 m/s
(a)
Given, horizontal force Fx = 45 N, vertical force Fy = 22 N
Thus, the total force acting on the is F = [(Fx)^2 + (Fy)2]^1/2
Therefore, F = [(45N)^2 + (22 N)2]^1/2 = 50.08 N
(b)
Constant velocity v = 22 m/s and time interval t = 3.0 s
The horizontal distance travelled by the is x = v t = (22 m/s) (3.0 s) = 66 m
The work done by the person pushing the hockey stick is W = F. x = (50.08 N) (66 m) = 3305.93 J
(c) The magnitude of the horizontal force is greater than (almost double) the vertical force, so the motion of is in the same direction as the horizontal force. The horizontal force dominates here.
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Find the direction and magnitude of the vectors (a) A = (25m)x + (-12m)y(b) B = (2.0m)x + (15m)y and(c) A + B
ANSWER
[tex]\begin{gathered} a)\theta=-25.6\degree;|A|=27.73m \\ b)\theta=82.4\degree;|B|=15.13m \\ c)\theta=6.3\degree^{};|A+B|=27.17m \end{gathered}[/tex]EXPLANATION
To find the direction of a two-dimensional vector, we apply the formula:
[tex]\theta=\tan ^{-1}(\frac{a_y}{a_x})[/tex]To find the magnitude of a two-dimensional vector, we apply the formula:
[tex]|a|=\sqrt[]{(a^{}_y)^2+(a_x)^2}[/tex]a) The direction of vector A is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{-12m}{25m}) \\ \theta=\tan ^{-1}(-0.48) \\ \theta=-25.6\degree \end{gathered}[/tex]The magnitude of vector A is:
[tex]\begin{gathered} |A|=\sqrt[]{(25m)^2+(-12m)^2} \\ |A|=\sqrt[]{625m^2+144m^2}=\sqrt[]{769m^2} \\ |A|=27.73m \end{gathered}[/tex]b) The direction of vector B is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{15m}{2m}) \\ \theta=\tan ^{-1}(7.5) \\ \theta=82.4\degree \end{gathered}[/tex]The magnitude of vector B is:
[tex]\begin{gathered} |B|=\sqrt[]{(2.0m)^2+(15m)^2} \\ |B|=\sqrt[]{4m^2+225m^2}=\sqrt[]{229m^2} \\ |B|=15.13m \end{gathered}[/tex]c) First, we have to find the sum of A and B:
[tex]\begin{gathered} A+B=(25m)x+(-12m)y+(2.0m)x+(15m)y \\ A+B=(25m+2.0m)x+(-12m+15m)y \\ A+B=(27m)x+(3m)y \end{gathered}[/tex]The direction of the vector (A + B) is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{3m}{27m}) \\ \theta=\tan ^{-1}(0.1111) \\ \theta=6.3\degree^{} \end{gathered}[/tex]The magnitude of the vector (A + B) is:
[tex]\begin{gathered} |A+B|=\sqrt[]{(27m)^2+(3m)^2} \\ |A+B|=\sqrt[]{729m^2+9m^2}=\sqrt[]{738m^2} \\ |A+B|=27.17m \end{gathered}[/tex]a student throws a rock straight down words from a bridge into a river below if initial speed of the rock is 10.0 Ms and it takes 2.1 s to reach the river how high is the bridge
Since this is a free fall and all the motion is going down, let's make down the positive direction (that way we won't use minus signs). A free fall is an uniform accelerated motion, this means that we can use the formula:
[tex]y-y_0=v_0t+\frac{1}{2}at^2[/tex]In this case we know the initial velocity is 10 m/s, the time is 2.1 and the acceleration of gravity is 9.8 m/s^2. Plugging this values we have that:
[tex]\begin{gathered} y-y_0=10(2.1)+\frac{1}{2}(9.8)(2.1)^2 \\ y-y_0=42.609 \end{gathered}[/tex]This means that the rock fell 42.609 meters. Therefore, the bridge is 42.609 m tall.
A block with a mass m1 is hit by a force of magnitude F which causes the block to have an acceleration of magnitude a. If a second block of mass m2 is hit by the same force of magnitude F which causes the block to have an acceleration of magnitude 2a, then which of these could be the two masses? A) m1= 200kg ; m2= 100kgB) m1= 50kg ; m2= 25 kgC) m1= 100kg ; m2= 50kgD)m1= 10kg ; m2= 50kgE)Any of these
We are given that a force "F" accelerates an object of mass "m1". According to Newton's second law, this can be represented by the following equation:
[tex]F=m_1a[/tex]Now, we are given that a second object of mass "m2" is accelerated by "2a" using the same force. Using Newton's second law we get:
[tex]F=m_2(2a)[/tex]Now, we will divide both equations, we get:
[tex]\frac{F}{F}=\frac{m_1a}{m_2(2a)}[/tex]Now, we simplify by canceling put the "F" and the "a":
[tex]1=\frac{m_1}{2m_2}[/tex]Now, we multiply both sides by "2m2", we get:
[tex]2m_2=m_1[/tex]Therefore, the first mass must be twice the second mass.
The options that meet this condition are:
[tex]m_1=200kg,m_2=100kg\text{ }[/tex][tex]m_1=50kg,m_2=25kg[/tex][tex]m_1=100kg,m_2=50kg[/tex]1) The net external force on a golf cart is 390 N north. If the cart has a total mass of 270 kg, what arethe magnitude and directions of its acceleration?
Given data
*The net external force on a golf cart is F = 390 N
*The cart has a total mass is m = 270 kg
The formula for the magnitude of the acceleration of the cart is given as
[tex]a=\frac{F}{m}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{390}{270} \\ =1.44m/s^2 \end{gathered}[/tex]Hence, the magnitude of the acceleration of the cart is 1.44 m/s^2. The direction of the acceleration is towards the north direction.
An object with a mass of 5kg is moving with a force of 25N. What is the object's acceleration?
According to Newton's second law, F=ma
Where F is the force acting on the object, m is the mass of the object, and a is the acceleration of the body due to the applied force.
It is given in the question,
m=5 kg
F=25 N
On substituting the known values in the above equation,
[tex]25=5\times a[/tex]On rearranging the above equation,
[tex]a=\frac{25}{5}=5m/s^2[/tex]Therefore the object's acceleration is 5 m/s²
A/An _____ is described as a device that detects current differences and then opens the circuit preventing electrocution.circuit breakerfuseground fault interruptershort circuit
ANSWER
ground fault interrupter
EXPLANATION
A ground fault interrupter is an automatic switch that measures the difference of current between the "hot" and neutral wires. If this difference is greater than a fixed values (only a few milliamperes) the switchs opens and stops the circulation or current.
If a person is in contact with a wire in the circuit the current is diverted through the person's body to the ground. Therefore the ground fault interrupter will detect a difference of current. Hence this device is used to prevent electrocution.
A park ranger driving on a back country road suddenly sees a deer in his headlights 20m ahead. The ranger, who is driving at 11.4 m/s, immediately applies the brakes andslows down with an acceleration of 3.80 m/s2. How much distance is required for theranger's vehicle to come to rest? Only enter the number, not the units,
Given data
*The speed of the ranger who is driving at 11.4 m/s
*The given acceleration is a = -3.80 m/s^2
The formula for the distance covered by the ranger's vehicle is given by the kinematic equation of motion as
[tex]\Delta x=\frac{v^2-v^2_0}{2a}[/tex]*Here v = 0 m/s is the initial speed of the ranger's vehicle
Substitute the values in the above expression as
[tex]\begin{gathered} \Delta x=\frac{0^2-(11.4)^2}{2\times(3.80)} \\ =17.1\text{ m} \end{gathered}[/tex]The distance is required for the ranger's vehicle to come to rest is calculated as
[tex]\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}[/tex]The stopping time is calculated as
[tex]v=v_0+at[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}[/tex]Homework: Action-reaction forces area. equal in magnitude and point in the same directionb. equal in magnitude and point in opposite directionsc. unequal in magnitude but point in the same directiond. unequal in magnitude and point in opposite directions
We will have that they are:
Equal in magnitude and point in opposite directions. {Option B]