[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]
Projectile is thrown with a velocity = v Angle of projection = θVelocity of projectile at a height half of the maximum height covered be [tex] \sf{v_0}[/tex][tex]\qquad[/tex]______________________________
Then –
[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]
[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]
[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]
[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]
Now, the vertical component of velocity of projectile at the height half of [tex] \sf{h_m}[/tex] will be –[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]
[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]
Therefore, the vertical component of velocity of projectile at this height will be–
☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]
Answer:
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2
What is the gravitational force between two objects of masses 20kg and 50kg separated by a distance of
4m?
Answer:
4.16875×10^-9
Explanation:
given,
mass(M1)=20kg
mass(M2)= 50kg
distance(d)=4m
gravitational constant value (G)=6.67×10^-11
we know,
gravitational force=(Gm1m2)÷d^2
=(6.67×10^-11×20×50)÷4^2
=(6.67×10^-8)÷16
= 4.16875×10^-9
A horizontal force F~ is applied to a block of mass m = 1 kg placed on an inclined
at θ = 30◦ plane. The coefficients of static and dynamic friction are µs = 0.3 and µ = 0.2, respectively. Find F such that the block is moving up the slope with a constant speed.
Hi there!
To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.
Recall:
[tex]\large\boxed{F_D = \mu N}}[/tex]
The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.
[tex]\large\boxed{N = Mgcos\theta + Fsin\theta}}[/tex]
We can plug in the known values to solve for one part of the normal force:
N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F
Now, we can plug this into the equation for the dynamic friction force:
Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F
For a block to move with constant speed, the summation of forces must be equivalent to 0 N.
If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:
Fcosθ = 1.697 + .1F
Solve for F:
Fcos(30) - .1F = 1.697
F(cos(30) - .1) = 1.697
F = 2.216 N
When the mass of the cylinder increased from 3. 0 kg to 6. 0 kg, what happened to the amount of heat generated in the system? It decreased by a factor of 3. It decreased by a factor of 2. It increased by a factor of 2. It increased by a factor of 3.
When the mass of the cylinder increases from 3. 0 kg to 6. 0 kg, the amount of heat generated in the system increases by a factor of 2.
The given parameters:
Initial mass of the cylinder, m₁ = 3.0 kgFinal mass of the cylinder, m₂ = 6.0 kgThe amount of heat generated in the system is calculated as follows;
Q = mcΔT
where;
m is the massc is the specific heat capacityΔT is the change in temperatureIf we keep every other parameters constant and change the mass from 3.0 kg to 6.0 kg, the increase in the heat generated is calculated as follows;
[tex]Q_1 = 3c \Delta T\\\\Q_2 = 6c\Delta T\\\\Q_2 = 2(3c \Delta T)\\\\Q_2 = 2Q_1[/tex]
Thus, when the mass of the cylinder increases from 3. 0 kg to 6. 0 kg, the amount of heat generated in the system increases by a factor of 2.
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define nuclear energy
Answer:
Nuclear energy is the energy stored in atoms that can produce electricity.
Hope that helps. x
SOMEONE, PLEASE HELP ME AS FAST AS YOU CAN, I WOULD GIVE MORE POINTS BUT THIS IS THE REST OF MY POINT, PLEASE SOMEONE WHO IS SUPER KIND HELP ME.
Answer:
1. The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time
2. 36÷4= 9
(formula) Speed = distance ÷ time
3. Meter (M) centimeter (cm) kilometers (km)
4. Seconds , Hour, Minute
5. 65÷13=5 (object A)
125÷ 25= 5 (object B)
their both Equal of the amount of speed, so they are travelling at the same speed
Hope I helped
A 1,500 kg car is traveling at 25 m/s. The driver suddenly applies the brakes causing the car to skids to a stop. If the average braking force between the tires and the road is 7,100 N, how far does the car slide before it comes to rest?
By Newton's second law, the car slows down with an average acceleration a such that
-7100 N = (1500 kg) a ⇒ a ≈ -4.7 m/s²
If we treat the car as accelerating uniformly with this magnitude, then the car slides a distance ∆x such that
0² - (25 m/s)² = 2a ∆x ⇒ ∆x ≈ 66 m
The car slide before it comes to rest will be 66 m. Newton's second equation of motion is applied in this problem.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Force is defined as the product of mass and acceleration. Its unit is Newton.
The acceleration is found as;
F=ma
-7100 N = (1500 kg) a
a= -4.7 m/s²
If we assume that the automobile accelerates uniformly with this magnitude, the car slides x distance. From the Newton's second equation of motion;
v²=u²+2as
v²-u²=2as
0² - (25 m/s)² =2×(-4.7) s
s=66 m
Hence,the car slide before it comes to rest will be 66 m.
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how do you think the liquid density affects the buoyancy of objects placed in the liquid?
Answer:
density can be written as:
d = F / V (weight density) or M / V (mass density)
Since F in the first equation can be written as
B = d V where B is the buoyant force and is proportional to the
weight of the liquid displaced
For a given volume the weight of the liquid displaced is proportional to the density and hence the buoyant force
is it possible for an object moving around a circular path to have both centripetal and tangential acceleration?
Answer:
A major difference between tangential acceleration and centripetal acceleration is their direction. Centripetal means “center seeking”. ... Tangential acceleration results from the change in magnitude of the tangential velocity of an object. An object can move in a circle and not have any tangential acceleration.
The direction of acceleration is a key distinction between tangential acceleration and centripetal acceleration. Centripetal is defined as "center-seeking." The amount of an object's tangential velocity changes, which causes tangential acceleration. Without experiencing any tangential acceleration, an item can move in a circle.
What is Acceleration?Acceleration is the rate at which the speed and orientation of a moving object change over time. When a spot or object moves faster or slower along a straight line, it is said to be accelerated. Even though the speed is constant, motion on a circle accelerates so because the direction is always shifting. Both by and to acceleration for all other types of motion.
It is a vector quantity because it has both magnitude and direction.
Any object's rotation experiences tangential acceleration, which is the speed at which the tangential velocity fluctuates. When an object is moving, it behaves in a tangential direction. An object moving in a circle will also experience the very same tangential velocity action.
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what is the necessary condition for the conservation of angular momentum
Answer:
The conservation of momentum of a system is a fundamental principle in classical mechanics. The only condition for this principle being valid is that the system should be an isolated one, i.e. It should not be acted upon by any external force.
What are the three different social perspectives on sport
Answer:
functionalist theory,feminist theory. discipline of sociology
☆ Correct and rewrite the following statements:
_____
e). electroplating is based on magnetic effect of electricity.
f). an electric bulb glows due to the chemical effect of electricity.
_______________________
#No spam
#No Internet answers, please #Explanation needed
Answer:
e)Question:-Electroplating is based on magnetic effect of electricity.Answer:-It's incorrect sentence As we know that, while electroplating there are two rods, one becoming cathode and another one becoming anode. During electroplating, when electricity is given through the rods, the anode produces positive ions which gets received at the cathode rod and negative ions produced at cathode collects at anode rod which results to the plating of the desired substance. Here, it is a chemical effect of electricity resulting to plating.So correct sentence is:
Electroplating is based of chemical effect of electricity. f)Question:-An electric bulb glows due to the chemical effect of electricity.Answer:-It's incorrect sentence As we know that, in an electric bulb there is a filament. When electricity is given in that filament of the electric bulb, the filament gets intensely heated up and results to the emission of light. Due to this heating of the filament due to electricity the electric bulb glows.So correct sentence is:
An electric bulb glows due to the heating effect of electricity.With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s
[tex]\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2[/tex]
[tex]\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2[/tex]
[tex]\\ \sf\longmapsto s=55+4(5.5)^2[/tex]
[tex]\\ \sf\longmapsto s=55+4(30.25)[/tex]
[tex]\\ \sf\longmapsto s=55+121[/tex]
[tex]\\ \sf\longmapsto s=176m[/tex]
what is the prmary source of energy inside of the earth
how to put in velocity into a graphing calculator
Answer:
You need a graphing calculator that implemented vectors. I don't know any that have it. Sorry.
Answer:
Provided an object traveled 500 meters in 3 minutes , to calculate the average velocity you should take the following steps:
Change minutes into seconds (so that the final result would be in meters per second). 3 minutes = 3 * 60 = 180 seconds ,
Divide the distance by time: velocity = 500 / 180 = 2.77 m/s .
Explanation:
A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hangs over a
frictionless pulley. When the mass is released, the cartaccelerates to the right
2.45 m/s2
4.90 m/s
9.80 m/s?
19.6 m/s
Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]
Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.
Let M = 1kg and m = 3 kg
Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.
When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,
M( g + a) = m(g - a)
1(9.8 + a) = 3( 9.8 - a)
9.8 +a = 29.4 - 3a
collect the like terms
4a = 19.6
a = 19.6/4
a = 4.9 m/[tex]s^{2}[/tex]
Therefore, the cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]
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Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.
Position (km) Time (h)
0.0 0.000
1.2 0.024
2.8 0.051
4.2 0.084
16.3 2.340
The average speed between 0 h and 2.340 h is 6.97 Km/h
Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.
[tex]Average \: speed = \frac{total \: distance}{total \: time} \\ \\ [/tex]
With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:
Total time = 2.340 – 0 = 2.340 hTotal distance = 16.3 – 0 = 16.3 KmAverage speed =?[tex]Average \: speed = \frac{total \: distance}{total \: time} \\ \\Average \: speed = \frac{16.3}{2.340} \\ \\ Average \: speed = 6.97 \: Km/h \\ \\ [/tex]
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Answer:
6.983 km/hr
Explanation:
The average rate of change is (16.34 - 0.0)/(2.34 - 0.000) ≈ 6.983 km/hr
Mejor yo por favor
Please HELP
6. A 3.4-kg bucket of water is attached to a 1.0-m rope. The bucket is swung in a circle at a speed
of 10.0-m/s.
a.
If the rope can only tolerate 400-N of force, what is the maximum speed the bucket can
experience before the rope snaps? {Hint: Let the centripetal force be 400-N and solve for the speed)
Answer:
Explanation:
hope this helps
The maximum speed of the bucket can experience before the rope snaps if A 3.4-kg bucket of water is attached to a 1.0-m rope. The bucket is swung in a circle at a speed of 10.0-m/s is 10.85 m / s.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the bucket, m = 3.4 kg,
The length of the rope, r = 1 m,
The speed of the bucket, v = 10 m / s,
The force of the rope, F = 400 N,
Calculate the maximum speed by the formula given below,
[tex]F_c = mv^2 / r[/tex]
400 = 3.4 v² / 1
v² = 400 / 3.4
v² = 117.6
v = 10.85 m / s
Therefore, the maximum speed of the bucket can experience before the rope snaps if A 3.4-kg bucket of water is attached to a 1.0-m rope. The bucket is swung in a circle at a speed of 10.0-m/s is 10.85 m / s.
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If a spaceship travels at an average speed of 6x10^10 kilometers per year, how many years will it take the spaceship to travel 3x10^30 kilometers? Show work or explain your logic for full credit. a) 5x10^2 b) 5x10^19 c) 5x10^20 d) 5x10^21
First, you need to find out the speed of 6x10^10. Then find out the value of 3x10^30 after you do that, Make this visual,
3x10^30. 5x10 to the 19 power
_______.
6x10^10
The answer is c
If the spaceship travels a distance of 6 × 10¹⁰ kilometers with an average speed the time taken to travel 3 × 10³⁰ kilometers is 5 × 10¹⁹ years. Thus, option a is correct.
What is speed?Speed is a physical quantity associated with a moving body. It is the measure of distance travelled per unit time . Thus, mathematically it is the ratio of distance to the time.
If we have the time and speed known we can find the distance travelled and vice versa. As the speed increases, the distance covered per unit time will be higher.
Here the spaceship travels 6 × 10¹⁰ kilometers . Thus the time taken to travel 3 × 10³⁰ kilometers can be calculated as follows:
1 year - 6 × 10¹⁰ km
? - 3 × 10³⁰ km
Cross multiply them we get, 3 × 10³⁰ km/ 6 × 10¹⁰ km = 5 × 10¹⁹ years.Hence, option a is correct.
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4) What process produces carbon dioxide?
a) photosynthesis
b) replication
c) mutation
d) respiration
5) What happens during cytokinesis?
a) a spindle forms
b) chloroplasts release energy
c) the cytoplasm divides
d) chromosomes
plz answer asap
Answer:
d and c are the answers :)
BRAINLIST!!!
Two metallic balls A and B carry the charges QA= 16×10^-9C and QB=-8×10^-9C.
We fixed balls A and B 5cm apart. Given e=1.6×10^-9C and k=9×10^9 S.I unit.
1) identify the ball that has excess of electrons.
2) calculate the number of deficit and excess of electrons on both balls.
3) calculate the force exerted by A on B.
the gravitational force acts on all objects in proportion to their mass. neglecting air resistance, why don’t heavy objects fall faster than light ones?
Answer:
Because in correspondence to the same distance from a mass, the gravitational acceleration is the same for all the bodies. It doesn't depend on the mass of the objects.
what is the theory relativity by Einstein?
:MAIN ANSWER:
determined that the laws of physics are the same for all non-accelerating observers
Explanation:
Albert Einstein, in his theory of special relativity, determined that the laws of physics are the same for all non-accelerating observers, and he showed that the speed of light within a vacuum is the same no matter the speed at which an observer travels, according to Wired.
HOPE IT HELPS
#CARRY ON LEARNING
HAPPY HOLIDAYS DEC 2021
Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.50 m. It takes the ball 0.70 seconds to complete one full circle. Calculate the tension in the string and its direction that provides the centripetal force acting on the ball to keep it in the circular path. (3 points)
0.60 N, along the line tangent to the circle
0.015 N, along the line tangent to the circle
0.60 N, toward the center of the circle
0.015 N, toward the center of the circle
Answer:
0.60N along the line tangent to the circle
Answer:
A is the answer
Explanation:
Just got finished with the quiz! Hope this helps <3
A water balloon is hovering directly above the line join points ANB which are 4.6 km apart if the angles of elevation to the balloon from point a to B or 28.8° and 52.2 respectively find the altitude of the balloon
Answer:
Drawing the triangle:
H / x = tan 52.2 = 1.29
H / (4.6 - x) = tan 28.8 = .550
H = 1.29 x
H = .55 * 4.6 - .55 x
1.84 x = 2.53 combining equations
x = 1.38
4.6 - 1.38 = 3.22
Total base of triangle = 1.38 + 3.22 = 4.6
H / x = tan 52,2 = 1.29
H = 1.29 * 1.38 = 1.78 height of triangle
Check:
1.78 / 3.22 = tan 28.9
This agrees with the given value of 28.8
what has to increase in order for an object to accelerate?
Answer:
Answer: B. Explanation: For an object to accelerate the force on it must be increased. According to Newton's second law of motion.
Explanation:
I do Accelerate to good luck
A rock is lying on a rock ledge that is 3 m high. The rock as 120 J of potential energy. What is the mass of the rock?
Have a great day!
[tex]\\ \sf\Rrightarrow PE=mgh[/tex]
[tex]\\ \sf\Rrightarrow m(10)(3)=120[/tex]
[tex]\\ \sf\Rrightarrow 30m=120[/tex]
[tex]\\ \sf\Rrightarrow m=4kg[/tex]
Mass=4kgRecoil is noticeable if you throw a heavy ball while standing on roller skates. If instead you go through the motions of throwing the ball but hold onto it, your net recoil velocity will be
Answer:
Since there is no external force, there is no change (movement) in the center of mass. Internally, the center of mass might change position, but the external result is still zero net velocity.
The net recoil velocity must be zero.
what is the cost of monthly (30 days) electric bill of ana if her city's cost of electricity is 0.05$ per kwh and she uses three refrigerators running in 600-watt power rating and open 24 hours
Answer:
1kW = 1000W
600W = 0.6kW
Cost of electric bill = 0.6kWh × 24 × 30 × $0.05
= $21.60
please help me !! i’ll give you brainliest if you get the answer right!
Answer: 25N to the right
Explanation: 35N - 10N = 25N
what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three signficant figures.
Answer:
Approximately [tex]5.11 \times 10^{-19}\; {\rm J}[/tex].
Explanation:
Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.
Look up the Rydberg constant for hydrogen: [tex]R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}[/tex].
Look up the speed of light in vacuum: [tex]c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}[/tex].
Look up Planck's constant: [tex]h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}[/tex].
Apply the Rydberg formula to find the wavelength [tex]\lambda[/tex] (in vacuum) of the photon in question:
[tex]\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}[/tex].
The frequency of that photon would be:
[tex]\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}[/tex].
Combine this expression with the Rydberg formula to find the frequency of this photon:
[tex]\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}[/tex].
Apply the Einstein-Planck equation to find the energy of this photon:
[tex]\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}[/tex].
(Rounded to three significant figures.)