True
Explanations:The position of an object can take any sign, it can be negative, zero or positive. This depends on the coordinate system chosen.
The position of an object (whether moving or static) is specified with respect to a frame of reference, and can be positive, zero or negative depending on the coordinate system chosen
Therefore, we can conclude that, in a chosen coordinate system, the position of an object in motion can have negative values.
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.
A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise as shown in the figure.
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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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
what is the general importance of water?
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.
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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Which of these is a property of an electromagnetic wave? A)magnetic and electric fields oscillate perpendicular to each other but not to the velocity of the wave B)transports energy C)has a magnetic wave but no electric wave
Electromagnetic waves :
- are transverse waves
- Can travel through a vacuum
- Tranport energy from one place to another
- can be reflected
-can be refracted
Correct options:
B)transports energy
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
Assume a water strider has a roughly circular foot of radius 0.0203 mm. The surface tension of water is 0.0700 N/m.A. What is the maximum possible upward force on the foot due to surface tension of the water? NB. What is the maximum mass of this water strider so that it can keep from breaking through the water surface? The strider has six legs. mg
Part (A)
The maximum possible upward force acting on the foot is,
[tex]F=2\pi r\sigma[/tex]Substitute the known values,
[tex]\begin{gathered} F=2(3.14)(0.0203\text{ mm)(}\frac{10^{-3}\text{ m}}{1\text{ mm}})(0.0700\text{ N/m)} \\ =8.9\times10^{-6}\text{ N} \end{gathered}[/tex]Thus, the maximum possible upward force on the foot is
[tex]8.9\times10^{-6}\text{ N}[/tex]Part (B)
The maximum force due to six legs can be expressed as,
[tex]6F=mg[/tex]Substitute the known values,
[tex]\begin{gathered} 6(8.9\times10^{-6}N)=m(9.8m/s^2) \\ m=\frac{6(8.9\times10^{-6}\text{ N)}}{9.8m/s^2}(\frac{1kgm/s^2}{1\text{ N}}) \\ =(5.45\times10^{-6}\text{ kg)(}\frac{1\text{ mg}}{10^{-6}\text{ kg}}) \\ =5.45\text{ mg} \end{gathered}[/tex]Thus, the maximum mass of water strider is 5.45 mg.
Neptune circles the Sun at a distance of 4.50 × 1012 m once every 164 years. Saturn circles the Sun at a distance of 1.43 × 1012 m. What is the orbital period of Saturn?
The orbital time period of the Saturn is 29.6 years
We are given that,
Distance from Sun to Saturn is = a = 1.43 × 10¹²
The mass of the Sun is = M =1.99 × 10³⁰kg
The Gravitational constant = G = 6.67 × 10⁻¹¹N-m²kg⁻²
To find the orbital period of Saturn we can use the equation ,
[tex]T^{2} = \frac{4\pi }{GM}a^{3}[/tex]
Where, T is the orbital time period of the of the Saturn , M is the mass of the sun , G is the gravitational constant.
Therefore, after putting the value in above equation we can get,
[tex]T^{2} = \frac{4(\(3.14)^{2} }{(6.67*10)^{-11} )N-m^{2} kg^{-2}}(1.43*10^{12}) ^{3}m[/tex]
[tex]T^{2} = \sqrt{8.688*10^{17} } s[/tex]
[tex]T = 932094415.818s[/tex]
So that , from above to convert the orbital time period of Saturn from second into year i.e. above seconds divided by seconds (1 sec = 3.154 ×10⁷ Earth years)
Thus, the orbital time period can be ,
[tex]T = 29.6 years[/tex]
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How much work is done on a medicine ball with a force of 29 newtons when you lift it 5 meters?
Given data
*The given force is F = 29 N
*The given distance is s = 5 m
The formula for the work is done on a medicine ball is given as
[tex]W=F\mathrm{}s[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} W=(29)(5) \\ =145\text{ J} \end{gathered}[/tex]Hence, the work is done on a medicine ball is W = 145 J
carts, bricks, and bands
10. Predict the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks.
a. Approximately 0.16 m/s2
b. Approximately 0.50 m/s2
c. Approximately 0.64 m/s2
d. Approximately 1.00 m/s2
D. The acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
What is acceleration?The acceleration of an object is the rate of change of velocity of the object with time.
The acceleration that would occur when four rubber bands were used to pull a cart loaded with two bricks, is determined by applying Newton's second law of motion as follows.
a = F/m
where;
a is accelerationF is the applied forcem is the massLet the mass of a brick = mass of a band = m
the mass of a cart = 2 bricks = 2m
a = (force applied by 4 rubber) / (mass of 1 cart + mass of 2 brick)
a = (4m) / (2m + 2m)
a = (4m)/(4m)
a = 1 m/s²
Thus, the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
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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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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.
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 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.
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.
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 train is traveling down a straight track at 26 m/s when the engineer applies the brakes, resulting in an acceleration of −1.0 m/s2 as long as the train is in motion. How far does the train move during a 52-s time interval starting at the instant the brakes are applied?______ m
In order to calculate the distance the train will move, we can use the formula below:
[tex]\Delta S=V_0t+\frac{at^2}{2}[/tex]Where V0 is the initial speed, t is the time and a is the acceleration.
Since the initial speed is 26 m/s and the train acceleration is -1 m/s², the train will completely stop after 26 seconds. In the remaining 26 seconds to complete the total of 52, the train will be stopped already, so there is no displacement.
Because of that, we will use a time t = 26 seconds.
So, using V0 = 26 m/s and a = -1 m/s², we have:
[tex]\begin{gathered} \Delta S=26\cdot26+\frac{(-1)26^2}{2}\\ \\ \Delta S=676-338\\ \\ \Delta S=338\text{ m} \end{gathered}[/tex]Therefore the train will move 338 meters.
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.
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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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
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
An object is placed 15cm in front of a convex mirror and an image is produced 5cm behind the mirror. Calculate the focal length of the mirror
The inverse of the optical power of an optical system, the focal length provides a measurement of how strongly the system converges or diverges light. A system's light converges when the focal length is positive, while it diverges when the focal length is negative.
A diverging mirror with the reflective surface bulging towards the direction of the light source is referred to as a convex mirror. Since they reflect light outward, they are not employed to focus light. Convex mirrors create an image that is smaller than the object and grows larger as it approaches closer to the mirror.
Given that the image of an object is generated v cm from a spherical mirror with a focal length of f, and that the object is situated u cm in front of the mirror, u, v, and f are connected by the equation 1/f=1/u + 1/v. The mirror formula is the term used to describe this equation. Both concave and convex mirrors can be used using the formula.
Therefore using, 1/f = 1/u + 1/v
1/f = 1/15 + 1/5
1/f = (1 + 3) / 15
f = 15/4 = 3.75cm
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Calculate the force between charges of 5.0x10^-8c and 1.0x10^-7c if they are .15m apart
Given data:
[tex]\begin{gathered} Q_1=5.0\times10^{-8}\text{ C} \\ Q_2=1.0\times10^{-7}\text{ C} \\ r=0.15\text{ m} \end{gathered}[/tex]The electrostatic force is given as,
[tex]F=\frac{KQ_1Q_2}{r^2}[/tex]Here, K is electrostatic force constant.
Substituting all known values,
[tex]\begin{gathered} F=\frac{9\times10^9\times5\times10^{-8}\times1\times10^{-7}^{}}{(0.15)^2} \\ =2\times10^{-3}\text{ N} \end{gathered}[/tex]Therefore, the force between the charges is 2×10^(-3) N. Hence, option (d) is the correct choice.
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
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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
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
