0.We are asked to determine the location of an image formed by an 7.75mm tall object that is located a distance of 17.5 cm from a convex mirror.
First, we will calculate the focal length using the following formula:
[tex]f=-\frac{R}{2}[/tex]Where:
[tex]\begin{gathered} f=\text{ focal length} \\ R=\text{ radius} \end{gathered}[/tex]Substituting the values we get:
[tex]f=-\frac{9.40cm}{2}[/tex]Solving the operations:
[tex]f=-4.7cm[/tex]Now, we use the following formula:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]Where:
[tex]\begin{gathered} d_0=\text{ distance of the object} \\ d_i=\text{ distance of the image} \end{gathered}[/tex]Now, we substitute the known values:
[tex]\frac{1}{17.5cm}+\frac{1}{d_i}=-\frac{1}{4.7cm}[/tex]Now, we solve for the distance of the image. First, we subtract 1/17.5 from both sides:
[tex]\frac{1}{d_i}=-\frac{1}{4.7cm}-\frac{1}{17.5cm}[/tex]Solving the operation:
[tex]\frac{1}{d_i}=-0.27\frac{1}{cm}[/tex]Now, we invert both sides:
[tex]d_i=\frac{1}{-0.27}cm=-3.7cm[/tex]Therefore. the location of the image is -3.7 centimeters.
The other parts are solved using the same procedure.
Part B. To calculate the size of the image we will use the following relationship:
[tex]\frac{h_i}{h_o}=-\frac{d_i}{d_0}[/tex]Where:
[tex]h_i,h_0=\text{ height of the image and height of the object}[/tex]Substituting we get:
[tex]\frac{h_i}{7.75mm}=-\frac{-3.7cm}{17.5cm}[/tex]Solving the operations on the right side:
[tex]\frac{h_i}{7.75mm}=0.21[/tex]Now, we multiply both sides by 7.75:
[tex]h_i=(7.75mm)(0.21)[/tex]Solving the operations:
[tex]h_i=1.64mm[/tex]Therefore, the height of the iamge is 1.64 mm.
The figure shows a person whose weight is W = 607 N doing push-ups. Find the
normal force exerted by the floor on (a) each hand and (b) each foot, assuming that
the person holds this position.
The normal force exerted by the floor on each hand is 203.95N and the normal force exerted on each foot is 99.55 N.
Torque is defined as force times the distance from the line of force that is perpendicular to it.
(a) Each hand's typical response is, let's say, N1. For hands, the overall typical response is 2N1.
Balance the torque around the foot now.
torque is applied in a counterclockwise direction
Now ,
2N1 x 1.25 = W x 0.84.
2N1 × 1.25 = 607 × 0.84
N1 = (607 x 0.84)/(2x1.25)
N1 = 203.95N.
(a) The normal response for each foot is, let's say, N2, and the total normal response for feet is 2N2.
Now, distribute torque among the hands.
Anticlockwise torque is equal to clockwise torque.
Now,
2N2 x 1.25 = W x 0.41
2N2 × 1.25 = 607 × 0.41
N2 = (607× 0.41 )/2× 1.25
N2 = 99.55N
Hence,the normal force exerted by the floor on each hand is 203.95N and the normal force exerted on each foot is 99.55 N.
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Please read the photo, Skill practice; how can you classify different types of energy?
There are 3 types of energy:
Kinetic energy(KE): energy of moving objects
Potential energy(PE): energy that an object has stored, position energy
Mechanical energy: The sum of both. PE+KE
For each, find the radius, diameter, and circumference of the circular object (one of these measurements is given in the problem). When working with the circumference, use π and round to the nearest whole number. a. Breaking a cookie in half creates a straight side 10 cm long.radius: cmdiameter: cmcircumference: cm
ANSWER
• radius: ,5 cm
,• diameter: ,10 cm
,• circumference: ,31 cm
EXPLANATION
If we assume that the cookie is round, when we cut it in half we'll have the following shape:
The straight side is the diameter of the cookie, which is 10 cm long.
The radius of a circle is half the diameter, hence the radius of the cookie is 5 cm.
The circumference is,
[tex]C=\pi\cdot d[/tex]Where d is the diameter of the circle. In this case, this is 10cm,
[tex]C=\pi\cdot10\operatorname{cm}\approx31\operatorname{cm}[/tex]The circumference of the cookie is 31 cm, rounded to the nearest whole number.
Given a DC battery of voltage, V = 4.00 V connected to a resistor R with a current I = 3.00 A through the resistor. What power is in this circuit? 15.5 W 12.0 W 39.4 W 45.5 W 8.88 W
12.0 W
Explanation
Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. to find the power in the circuit we need to use the expression:
[tex]P=IV[/tex]where P is the powe I is the current and V is the voltage
Step 1
a)Let
[tex]\begin{gathered} I=\text{ 3 Amperes} \\ V=4.0\text{ volts} \end{gathered}[/tex]b) now,replace
[tex]\begin{gathered} P=IV \\ P=3\text{ A*4 V} \\ P=12\text{ W} \end{gathered}[/tex]therefore, the answer is
12.0 W
I hope this helps you
A block of mass 0.500 kg slides on a flat smooth surface with a speed of 2.80 m/s. It then slides over a rough surface with μk and slows to a halt. While the block is slowing, (a) What is the frictional force on the block? (b) What is the magnitude of the block’s acceleration? (c) How far does the block slide on the rough part before it comes to a halt?
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
m = 0.5 kg
v = 2.8 m / s
Since there is no vertical motion,
∑ [tex]F_{y}[/tex] = 0
N - mg = 0
N = 0.5 * 9.8
N = 4.9 N
μ = 0.3
[tex]f_{k}[/tex] = μ N
[tex]f_{k}[/tex] = 0.3 * 4.9
[tex]f_{k}[/tex] = 1.47 N
The net force acting on the block is due to friction,
F = [tex]f_{k}[/tex] = 1.47 N
F = m a
1.47 = 0.5 * a
a = 2.94 m / s²
Since, acceleration is towards the opposite of motion,
a = - 2.94 m / s²
v² = u² + 2 a s
0 = 2.8² + ( 2 * - 2.94 * s )
s = 1.33 m
Therefore,
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
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An unbanked asphalt highway has turns of 40m radii. How fast should the speed limit be if cars may be traveling in the rain? (Us for wet asphalt on rubber is .755)
Given data:
* The radius of the turn is r = 40 m.
* The coefficient of friction is,
[tex]\mu_s=0.755[/tex]Solution:
The centripetal force acting on the car is,
[tex]F=\frac{mv^2}{r}[/tex]where m is the mass of the car,
The frictional force acting on the car is,
[tex]F_r=\mu_smg[/tex]where g is the acceleration due to gravity,
In order to travel a car in rain, the centripetal force acting on the car must be equal to the frictional force on the same car.
Thus,
[tex]\begin{gathered} F_r=F_{} \\ \mu_smg=\frac{mv^2}{r} \\ \mu_sg=\frac{v^2}{r} \\ v^2=r\mu_sg \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} v^2=0.755\times40\times10 \\ v^2=302 \\ v=17.4\text{ m/s} \end{gathered}[/tex]Thus, the maximum speed limit of the car in rain is 17.4 m/s.
Hence, the nearest possible correct answer is option b.
An archery bow is drawn a distance d = 0.39 m and loaded with an arrow of mass m = 0.088 kg. The bow acts as a spring with a spring constant of k = 195 N/m, and the arrow flies with negligible air resistance. To simplify your work, let the gravitational potential energy be zero at the initial height of the arrow. If the arrow is shot at an angle of θ = 45° above the horizontal, how high, in meters above the initial height, will the arrow be when it reaches its peak?
The maximum height reached by the arrows is determined as 8.6 m.
What is the initial speed of the arrow?The initial velocity of the arrow is calculated by applying the principle of conservation of energy as shown below;
K.E = U
where;
K.E is the kinetic energy of the arrowU is the elastic potential energy of the bow¹/₂mv² = ¹/₂kx²
mv² = kx²
v² = kx²/m
v = √(kx²/m)
where;
k is spring constant of the bowm is the mass of the arrowx is the extension of the bowv = √(195 x 0.39²/0.088)
v = 18.36 m/s
The maximum height reached by the arrow is calculated as follows;
H = (v² sin²θ) / (2g)
where;
θ is angle of projection of the arrowg is acceleration due to gravityH = (18.36² (sin45)²) / (2 x 9.8)
H = 8.6 m
Thus, the height of the arrow above the ground when it reaches its peak is 8.6 m.
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if i kicked a empty soda can would it travel further than a filled up soda can, if so why? & what newton law would this be ?
If we kick both sodas with the same force, the soda that has a higher weight will have a lower acceleration. This is explained by Newton's second law of motion
The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
F = ma
where
m = mass
a = accelaration
F = Force
In which direction does the tension force on the pirate point when a gust of wind rises up?
Answer: For downwind sailing, with the sail oriented perpendicular to the wind direction, the pressure increase on the upwind side is greater than the pressure decrease.
Explanation: this is all i know
How much kinetic energy does Usain Bolt (m=94kg) have when he hits his top
speed of 12 m/s?
Answer:
6768 Joules (J)
Explanation:
kinetic energy = 1/2mv^2
1/2 (94x12^2) = 6768
Which label goes through the horizontal axis?
A. Distance
B. Acceleration
C. Force
D. Mass
Answer:
A: Distance
Explanation:
You can not change gravity with acceleration or force in a weird way. Only mass and distance really affect gravity in normal physics.
Gravity is stronger when two objects are closer. Therefore gravity would decrease as distance increased.
Mass increases gravity.
The image below shows that gravity (depth) increases as the distance from the black hole decreases.
Two cars are driving at 25 mph and one of them hits a brick wall and stops immediately. The other one runs into a pile of old beanbag chairs and slowly comes to a stop. Which one has the greater
ANSWER
The one that hits the wall.
EXPLANATION
We want to identify the situation in which there is a greater impulse.
The impulse of an object is given as a change in its momentum.
Mathematically, it is given as:
[tex]\Delta p=F\Delta t[/tex]where F = force
Δt = change in time
As we can see, impulse depends on the impact force and the amount of time that the force is applied.
This means that if a collision takes a longer time to occur, the force acting on the object is much smaller compared to a situation where the collision takes a short time to occur.
If the force acting on the object is smaller, it, therefore, implies that the impulse will also be lesser.
Hence, the car that hits the wall will have a greater impulse than the one that hits the beanbag chairs, since that collision took a longer time to occur.
The weight of a proton is 1.64×10−26 N. The charge on a proton is +1.60×10−19 C. If a proton is placed in a uniform electric field so that the electric force on the proton just balances its weight, what is the magnitude and direction of the field?
Given:
The weight of the proton is: W = 1.6 × 10^(-26) N.
The charge on a proton is: q = 1.60 × 10^(-19) C
To find:
The magnitude and the direction of the electric field.
Explanation:
The weight of the proton is the force that the proton experiences due to its mass and acceleration. The electric force balances the weight of the proton. Thus we have,
F = W
Here, F is the electric force a proton experiences when it is placed in an electric field and W is the weight of the proton,
The force experienced by a photon when it is placed in an electric field is given as,
[tex]F=Eq[/tex]Here, E is the electric field.
Rearranging the above equation and substituting the values, we get:
[tex]\begin{gathered} E=\frac{F}{q} \\ \\ E=\frac{1.64\times10^{-26}\text{ N}}{1.60\times10^{-19}\text{ C}} \\ \\ E=1.025\times10^{-7}\text{ N/C} \end{gathered}[/tex]Thus, the magnitude of the electric field is 1.025 × 10^(-7) N/C.
The charge on the proton is positive and when it is placed in the electric field, the electric force on the proton is balanced by the weight of the proton. Thus, The direction of the electric force is opposite to the direction of the weight of the proton which is radially outward.
Final answer:
The magnitude of the electric field is 1.025 × 10^(-7) N/C and it has a radially outward direction that is opposite to the wight of the proton.
The safe loadof a wooden beam supported at both ends varies jointly as the width, w, the square of the depthd, and inversely as the length A wooden beam 7 inwide, 10 indeepand 19 ft long holds up 4422 What load would a beam inwide. 5 indeep and long of the same material support? (Round off your answer to the nearest pound )
L = k(wd^2/ l)
L = load
L1= 4422 lb
w = width
w1= 7 in
d = depth
d1= 10 in
l = lenght
l1= 19 ft
L2=
w2= 5in
d2=5in
l2= 11ft
First solve the constant with values 1
L1= k [(w1) (d1)^2 / l1]
4422= k [(7) (10)^2 / 19]
k = 4422 / [(7) (10)^2 / 19]
k= 120 lb/ft in^3
Replace with values 2
L2= k [(w2) (d2)^2 / l2]
L2= 120 [(5) (5)^2 / 11]
L2= 1363.63 LB
A dentist causes the bit of a high speed drill to accelerate from an angular speed of 1.76 x 10^4 rads to an angular speed of 4.61 x 10^4 rat. In the process, the bit turns through 1.97 x 10 ^4 rad. Assuming a constant angular acceleration, how long would it take the reach its maximum speed of 7.99 x 10^4 rads starting from rest?
The time taken for the bit to reach the maximum speed is 1.35 seconds.
What is the angular acceleration of the bit?The angular acceleration of the bit is determined by applying the following kinematic equation as shown below.
ωf² = ωi² + 2αθ
where;
ωf is the final angular speedωi is the initial angular speedθ is the angular displacementα is the angular accelerationα = (ωf² - ωi²)/2θ
α = (46,100² - 17,600²) / (2 x 19,700)
α = 46,077.4 rad/s²
The time taken for the bit to reach the maximum speed is calculated as follows;
ωf = ωi + αt
t = (ωf - ωi) / α
t = (79,900 - 17,600) / (46,077.4)
t = 1.35 seconds
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i’m still really confused on how to actually calculate it
Question 6:
Given information:
Distance travelled by bus,
[tex]s=10100\text{ m}[/tex]Average velocity of the bus,
[tex]v=5.6\text{ m/s}[/tex]We need to find the time taken by bus to reach school. Let t be the time taken by bus to reach school. The velocity of the bus is given as,
[tex]v=\frac{s}{t}[/tex]The expression for the time is given as,
[tex]t=\frac{s}{v}[/tex]Substituting all known values,
[tex]\begin{gathered} t=\frac{10100\text{ m}}{5.6\text{ m/s}} \\ \approx1804\text{ s} \\ \approx30\text{ min 4 sec} \end{gathered}[/tex]Therefore, the bus required 30 min 4 sec to reach school.
The mass of the Moon is about 1/80th of the mass of Earth. The force exerted by Earth on the Moon is about 80 times thatexerted by the Moon on Earth.Select one:O TrueO False
According to Newton's Third Law of Motion, the force that object A exerts to object B has the same magnitude as the force that object B to object A, but in the opposite direction:
[tex]\vec{F}_{AB}=-\vec{F}_{BA}[/tex]Then, the force exerted by Earth on the Moon has the same magnitude as the force exerted by the Moon on the Earth.
Therefore, the given statement is false.
a projectile starting from ground hits a target on the ground located at the distance of 1000m after 40 sec
A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) The size of the angle θ is = 83°
b)The initial velocity was the projectile launched is (u)=205.13 m/s.
What is velocity?The velocity is a physical term that is refer to how much the object has covered the distance in a given time. It can be measured in m/s and cm/s.
How can we calculate the velocity?a) To calculate the angle we are using two formulas, they are
r= u cos(θ)T
T= 2u sin (θ)/g
Here we are given,
r= The distance covered in the motion = 1000m.
T = The time covered in the motion = 40 Seconds.
g= The acceleration due to gravity = 9.8 m/s²
We have calculate the values of angle = θ
Now we put the values in above equation we get,
1000= u cosθ*40...(1)
40*g= 2u sinθ.....(2)
Equation(2) divided by equation(1) we get,
2tanθ=40*40*g/1000
Or, tanθ=7.84
Or, θ= 82.7°≈83°
From the above calculation we can say that, The size of the angle θ is = 83°
b) The initial velocity was the projectile launched is = u
As we know, r= u cos(θ)T
Now we put the values in the equation we get,
1000= u cos(83)*40
Or, u=205.13 m/s
From the above calculation we can say that, The initial velocity was the projectile launched is (u)=205.13 m/s.
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A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) What is the size of the angle θ?
b) At what initial velocity was the projectile launched?
Which of the following statements about air is TRUE?
A. Air is not a source of resistance.
B. Air has mass, but not inertia.
C. Air is not affected by human movement.
D. None of these statements are true.
The true statement among the following is that the air is not affected by human movement. Hence, option C is correct.
What is Air?Air relates to the atmosphere of the planet. Several gases and minute dust particles make up the air. Living organisms breathe and thrive in this pure gas. Its shape and volume are ill-defined. Considering that it is matter, it has mass and weight. Atmospheric pressure is generated by air weight. The space vacuum lacks air.
About 78% of the gas within air is nitrogen, 21% of the gas is oxygen, 0.9% of the gas is argon, 0.04% of the gas is co2, and very little other gas is present.
A typical amount of water vapor is around 1%.
Since respiration requires oxygen, animals must breathe it to survive. The lungs transfer back carbon dioxide back into the atmosphere when breathing, putting oxygen into the body.
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Determine the resistance, in milliOhms, of a metal rod 2.96 m long, 0.89cm diameter and composed of aluminum of resistivity 2.8 x 10-8 Ωm .
The resistance R of a rod with length L, cross-sectional area A and resistivity ρ is given by:
[tex]R=\frac{\rho L}{A}[/tex]On the other hand, the area of a circle with diameter D is given by:
[tex]A=\frac{\pi}{4}D^2[/tex]Then, the resistivity of the rod in terms of its diameter is:
[tex]R=\frac{4\rho L}{\pi D^2}[/tex]Replace L=2.96m, D=0.89cm and ρ=2.8×10^(-8)Ωm to find the resistance of the metal rod:
[tex]\begin{gathered} R=\frac{4\rho L}{\pi D^2} \\ \\ =\frac{4(2.8\times10^{-8}\Omega m)(2.96m)}{\pi(0.89cm)^2} \\ \\ =\frac{4(2.8\times10^{-8}\Omega m)(2.96m)}{\pi(0.89\times10^{-2}m)^2} \\ \\ =1.332232...\times10^{-3}\Omega \\ \\ \approx1.33m\Omega \end{gathered}[/tex]Therefore, the resistance of the metal rod is approximately 1.33 miliOhms.
list advantages and disadvantages of surface tension
please i really need this urgently
Advantages of Surface tension are -
It gathers water into a ball.
It permits a distinct boundary layer that is similar to a non-Newtonian liquid.
It enables water to rise into a paintbrush through capillary action.
It permits rain to fall as drops as opposed to a stifling mass.
It enables smooth surfaces to form as concrete and liquid metals solidify.
Disadvantages of Surface tension are -
The behavior of water would alter if surface tension were eliminated, some of these changes being related to surface tension's drawbacks. Washing clothing is one example that is close to home. Detergent is required while washing garments due to the comparatively high surface tension of water. Reduced surface tension enables laundry water to fully permeate the garments for better cleaning as part of the task of laundry detergent. It would take far less detergent to wash clothes if water had a naturally low surface tension.
When you want to create a fine water spray, such as with a lawn sprinkler, surface tension again becomes an issue. Surface tension makes it harder to divide water into tiny droplets. Sprinklers could operate on a hose with less pressure if they were used with water that had a lower surface tension.
What is a surface tension ?
Surface tension is the propensity for liquid surfaces that are at rest to condense into the smallest surface area. Razor blades and insects (like water striders), which have a higher density than water, can float on the surface of the water without even becoming partially buried because to surface tension.
Surface tension at liquid-air contacts originates from the liquid molecules' stronger attraction to one another due to cohesion than to the air molecules (due to adhesion).
There are primarily two mechanisms at work. One causes the liquid to constrict by exerting an inward push on the surface molecules. The second force is tangential and parallel to the liquid's surface. The surface tension is the common name for this tangential force.
Overall, the liquid acts as though an elastic membrane was stretched over its surface. However, this comparison should not be drawn too far because surface tension is a characteristic of liquid-air or liquid-vapor interfaces, but the tension in an elastic membrane depends on how much it is deformed.
Advantages of Surface tension are -
It gathers water into a ball.
It permits a distinct boundary layer that is similar to a non-Newtonian liquid.
It enables water to rise into a paintbrush through capillary action.
It permits rain to fall as drops as opposed to a stifling mass.
It enables smooth surfaces to form as concrete and liquid metals solidify.
Disadvantages of Surface tension are -
The behavior of water would alter if surface tension were eliminated, some of these changes being related to surface tension's drawbacks. Washing clothing is one example that is close to home. Detergent is required while washing garments due to the comparatively high surface tension of water. Reduced surface tension enables laundry water to fully permeate the garments for better cleaning as part of the task of laundry detergent. It would take far less detergent to wash clothes if water had a naturally low surface tension.
When you want to create a fine water spray, such as with a lawn sprinkler, surface tension again becomes an issue. Surface tension makes it harder to divide water into tiny droplets. Sprinklers could operate on a hose with less pressure if they were used with water that had a lower surface tension.
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Jeff tosses a can of soda pop to Karen, who is standing on her 3rd floor balcony a distance of 8.5m above Jeff’s hand. Jeff gives the can an initial velocity of 16m/s, fast enough so that the can goes up past Karen, who catches the can on its way down. Calculate the velocity of the can the instant before Karen grabs the can. How long after Jeff tosses the can does Karen have to prepare to catch it?
ANSWER
9.51 m/s
EXPLANATION
We know that Jeff is 8.5m below Karen. He tosses the can up with initial velocity u = 16m/s and it passes where Karen is, so the maximum height of the can is 8.5m plus some more meters x. Then Karen catches the can in its way down, so when she does the can goes this distance x.
Let's find this distance. The height of an object thrown up with initial velocity u is:
[tex]y=ut-\frac{1}{2}gt^2[/tex]We know u = 16m/s but we don't know the time. This we can find from the final velocity of the can:
[tex]v=u-gt[/tex]At its maximum height the velocity is zero:
[tex]0=u-gt[/tex]Solving for t:
[tex]t=\frac{u}{g}[/tex]If we assume g = 9.8m/s²:
[tex]t=\frac{16m/s}{9.8m/s^2}=1.63s[/tex]We know that the can was in the air for 1.63 seconds until it reached its maximum height. The maximum height is:
[tex]y=16m/s\cdot1.63s-\frac{1}{2}\cdot9.8m/s^2\cdot1.63^2s^2[/tex][tex]y=26.08m-13.02m=13.06m[/tex]This is the maximum height of the can. The extra distance the can travelled above Karen is:
[tex]x=13.06m-8.5m=4.56m[/tex]In the can's way down, the initial velocity is 0, because it starts falling after stopping in its way up. The acceleration is still the acceleration of gravity and the height it falls is x. We can find the time it took to reach Karen's hand after it started falling:
[tex]y=\frac{1}{2}gt^2[/tex]Note that in this case we use the acceleration of gravity positive because it is in the same direction of the can's motion. Solving for t:
[tex]t=\sqrt[]{\frac{2y}{g}}[/tex][tex]t=\sqrt[]{\frac{2\cdot4.56m}{9.8m/s^2}}=\sqrt[]{0.93s^2}=0.97s[/tex]Knowing that the can was in the air for another 0.97 seconds after starting falling until it reached Karen's hand, we can find its velocity at that instant:
[tex]v=u+gt[/tex]Remember that in this case u = 0:
[tex]v=gt=9.8m/s^2\cdot0.97s=9.51m/s[/tex]The velocity of the can the instant before Karen grabs it is 9.51 m/s
How many moles of a gas sample are in a 5.0 L container at 251 K and 370 kPa?(The gas constant is8.31L kPamol K)Round your answer to one decimal place and enter the number only with no units.
Given:
The volume of the gas, V=5.0 L
The temperature of the gas, T=251 K
The pressure of the gas, P=370 kPa
The gas constant, R=8.31 L kPa/(mol K)
To find:
The moles of the gas sample.
Explanation:
From the ideal gas equation,
[tex]PV=\text{nRT}[/tex]Where n is the moles of the gas present.
On substituting the known values,
[tex]\begin{gathered} 370\times5.0=n8.31\times251 \\ \Rightarrow n=\frac{370\times5.0}{8.31\times251} \\ =0.9\text{ mol} \end{gathered}[/tex]Final answer:
The moles of the gas present in the sample is 0.9 mol
Explain why clothes stick together when they are removed from a drier. What is static electricity?
ANSWER:
What happens in clothes is a phenomenon called static cling is a phenomenon caused by static electricity. When dry materials rub against each other, they can exchange electrons, creating an electrical charge. This charge can build up in the form of static electricity and cause two objects, in this case clothing, to stick or stick together.
When the substance that loses electrons becomes positively charged and the substance that gains electrons becomes negatively charged. These charges are stationary and remain on the surface of the material. Since there is no flow of electrons, this is called static electricity.
The owner of a recycling company wants to reduce his electrical consumption and costs. The electromagnet used in his operation uses 12 A of current, has 7000 loops and a lifting force of 9800 N. If the lifting force needs to remain the same but the owner would like to reduce the current to only 5 A, how many loops would the electromagnet have?
Given:
Current, I = 12 A
Loops, B = 7000
Force, F = 9800 N
Let's determine the loops if the force remains the same but the current redudces to 5A.
Apply the formula:
[tex]F=\frac{I\times N}{L}[/tex]Let's solve for L.
[tex]\begin{gathered} L=\frac{I\times N}{F} \\ \\ L=\frac{12\times7000}{9800} \\ \\ L=8.57\text{ m} \end{gathered}[/tex]If the current reduces to 5 A, we have:
[tex]\begin{gathered} N=\frac{F\times L}{I} \\ \\ \text{Where I = 5 A} \\ \\ N=\frac{9800\times8.57}{5} \\ \\ N=16800\text{ } \end{gathered}[/tex]The number of loops the electromagnet would have is 16800 loops.
Before they were decommissioned, the NASA space shuttles required two solid rocket boosters (SRBs) to launch the shuttle from Earth’s surface. Both SRBs produced 1.7 x10^7 N at liftoff. The combined mass of a shuttle and rocket boosters was about 1.5 x 10^ 6 kga) Calculate the net acceleration of a space shuttle and rockets at the time of liftoff. (b) Calculate the speed of the shuttle and rockets after 10.0 s.
Given:
The force on the booster is
[tex]F=\text{ 1.7}\times10^7\text{ N}[/tex]The mass is
[tex]m=\text{ 1.5}\times10^6\text{ kg}[/tex]Required:
(a) The net acceleration
(b) Speed of the shuttle and rockets after time t = 10 s
Explanation:
(a) The net acceleration can be calculated as
[tex]\begin{gathered} a=\frac{F}{m} \\ =\frac{1.7\times10^7}{1.5\times10^6} \\ =11.33\text{ m/s}^2 \end{gathered}[/tex](b)
The initial speed of the rocket and shuttle will be zero.
The speed of the rocket and shuttle after time t = 10 s will be
[tex]\begin{gathered} v=0\text{ m/s + 11.33}\times10 \\ =\text{ 113.3 m/s} \end{gathered}[/tex]Final Answer:
(a) The net acceleration is 11.33 m/s^2.
(b) The speed of the rocket and shuttle is 113.3 m/s
A driver of a car going 90km/hr suddenly sees the lights of a barrier 40.0m ahead. It take the driver 0.75s before he applies the brakes (this is known as reaction time). Once he does begin to brake, he decelerates at a rate of 10m/s^2. Does he hit the barrier?
First, consider that the distance traveled by the car in 0.75s is:
[tex]x=v\cdot t[/tex]Convert 90km/h to m/s as follow:
[tex]\frac{90\operatorname{km}}{h}\cdot\frac{1h}{3600}\cdot\frac{1000m}{1\operatorname{km}}=\frac{25m}{s}[/tex]Then, the distance x is:
[tex]x=(\frac{25m}{s})(0.75s)=18.75m[/tex]Then, when the driver start to apply the brakes, the distance to the barrier is:
x' = 40.0 m - 18.75 m = 21.25 m
Next, calculate the distance that the car need to stop completely, by using the following formula:
[tex]v^2=v^2_o-2ad[/tex]where,
v: final velocity = 0m/s (the car stops)
vo: initial velocity = 25m/s
a: acceleration = 10m/s^2
d: distance = ?
Solve the previous equation for d and replace the values of the other parameters:
[tex]d=\frac{v^2_0-v^2}{2a}=\frac{(\frac{25m}{s})^2-(\frac{0m}{s})^2}{2(\frac{10m}{s})^{}}=31.25m[/tex]Then, the drive needs 31.25 m to stop. If you compare the previous result with the distance of the car related to the barrier when the driver applies the brakes
(x' = 18.75 m), you can notice that d is greater than x'.
Hence, the car does hit the barrier.
The virtual image as seen in a plane mirror is reversed both left-to-right and top-to-bottom. Is this true or false?
ANSWER:
false
STEP-BY-STEP EXPLANATION:
The virtual images in a plane mirror have a left-right investment. But not top-to-bottom, which means that what the statement says is false.
jerome pitches a baseball of mass 0.300 kg. the ball arrives at home plate with a speed of 40.0 m/s and is batted straight back to jerome with a return speed of 52.0 m/s. what is the magnitude of change in the ball's momentum?
The change in the momentum of the ball is 3.6 Kgm/s.
What is momentum?The term momentum has to do with the product in the velocity of a body and mass of the body. We have to recall at this point that rate of change of momentum is directly related to the impressed force and this is in accordance with the Newton second law of motion.
Now we have to look at the few pieces of information that we can be able to glean from the question;
Mass of the object = 0.300 kg
Initial speed of the object = 40.0 m/s
Final speed of the object = 52.0 m/s
Given that the change in the velocity of the object is given by;
m( v - u)
m = Mass of the object
v = Final speed of the object
u = Initial speed of the object
change in the velocity of the object = 0.300(52 - 40)
= 3.6 Kgm/s
Learn more about momentum:https://brainly.com/question/904448
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IBangkok, Thailand, and Lima, Peru, are located at opposite positions of Earth's surface. When a rock is dropped in both cities, observers would describe the motion of either rock with a word that means "down."
Where are the two rocks moving?
Responses
They are moving away from each other.
.
One rock is moving due east and the other is moving due west.
They are moving down, which is the same direction in both cities.
They are moving toward each other.
Answer :dawg you got snaids thats the answer awww man
Explanation:
answer is that of an answer
