a. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by how much is the spring is compressed.
b. A 0.010 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position.
c. The same dart is now fired horizontally from a height of 4.70 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.
d. Find the horizontal distance from the equilibrium position at which the dart hits the ground.

Answer:

Explanation:

The formula for Speed is distance per time.

Data

D = 40 m

T = 10 sec

S = ?

To find the speed , divide the distance by the time.

Note : the unit for Speed is meter /second

Therefore the speed of the wave, the surfer is riding on is 4 m/s.

When a ball is thrown up by doing work, the velocity of the ball will be 2.6 m/s.

It states that the Work done in moving a body is equal to the change in kinetic energy of the body

Kinetic energy = 1/2 mv²

Given is a ball of mass m = 0.750 kg and the work done on ball W = 2.50 J

The ball is initially at rest. So, initial velocity is zero. Then, change in kinetic energy will be

W= ΔK.E = K.Ef - K.Ei

According to work energy theorem, work done is

W = 2.5J  = 1/2 x 0.750 x (v)² -0

v =2.6 m/s

Thus, the velocity of the ball is 2.6 m/s

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NGC 300 is different from our solar system because of its smaller size.

NGC 300 galaxy is different from our solar system because of its size. NGC 300 is smaller than our solar system which means that it has less number of stars as compared to our solar system.

So we can conclude that NGC 300 is different from our solar system because of its smaller size.

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Bobby and Alice are pushing a stalled car in the same direction. The mass of the car is 2,000 kg. Bobby applies 400 N to the car while Alice applies 300 N. A force created by friction is 500 N in the opposite direction is the acceleration of the car is 0.1 m/sec.

An acceleration, fee at which pace adjustments with time, in phrases of each velocity and direction. A factor or an item shifting in a direct line is expanded if it hurries up or slows down.

Is summarized via way of means of the equation: Force (N) = mass (kg) × acceleration (m/s²).The mass = 2000 kg and force= 1200 N. So mass / force = 2000/1200 = 0.1 m/sec.

Thus, the acceleration of the car s 0.1 m/sec.

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Answer:

coplanar When all forces are acting in the same 

resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition

The acceleration will be constantly equal to 1 m/s².Option c is correct.

Force is defined as the push or pulls 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.

Given data;

Force,F = 350 N

Friction force,f=350 N

F=f

Because the frictional and applied forces on the two boxes are of equal strength, both move at the same pace. As a result, the acceleration will be constantly equal to 1 m/s².

Hence option c is correct.

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(a) The initial speed must the object be launched so that it reaches

the surface of Planet 2 with zero speed is  √[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}]

(b) An inequality between M₁ and M₂ that represents when (a)

can occur is { M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0

(c)  if R₁ = R₂, then M₁ must be greater than M₂ is proved.

The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.

The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.

F = GMm/R²

Given, Planet 1 has mass M₁ and radius R₁. Planet 2 has mass M₂ and radius R₂. The two planets are a distance of L apart, measured from surface to surface. An object is launched with some initial speed from the surface of Planet 1 directly towards Planet 2. For this problem, assume that Planets1 and 2 are stationary.

If v is the launch velocity, then initial total energy will be

T.E  = 1/2 mv² + ( -GM₁m/ R₁ - G M₂m/(R₂ + L)

The final total energy will be

T.E  =0 + ( -GM₁m/ (R₁ +L) - G M₂m/ R₂)

From energy conservation principle, we get

1/2 mv² + ( -GM₁m/ R₁ - G M₂m/(R₂ + L) = ( -GM₁m/ (R₁ +L) - G M₂m/ R₂)

v = √[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}]

(b) an inequality between M₁ and M₂  so that object reaches the surface of Planet 2 with zero speed is

[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}] =0

{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0

Thus, this is an inequality between M₁ and M₂.

(c) If R₁ = R₂, then

{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0

M₁(R+L) + M₂R - M₁R  -M₂(R+L) / R (R+L)   ≥ 0

M₁(R+L) + M₂R - M₁R  -M₂(R+L)  ≥ 0

M₁L - M₂L  ≥ 0

M₁ ≥ M₂

M₁ must be greater than M₂.

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Answer:

A Rechargeable Battery

Explanation:

That is roughly what happens when a secondary cell is recharged (rechargeable battery.) Electric current transforms one chemical species with a lower energy level into another with a higher energy level. When the battery is loaded, this energy can be released. Strictly speaking, no conversion occurs: chemical energy is electrical energy! The energy of electrons in any particular substance, such as metal, are due to electron interactions with the nuclei in the substance and with each other. Electrons, in particular, occupy certain energy levels in metals. Because each metal has a distinct highest occupied energy level, electrons can alter their energies by traveling from one metal to another. An electrical current is made up of this movement!

Answer:

As electrical power is measured in Watts (W), therefore it must be also be measured in Joules per Second. So we can correctly say that: 1 watt = 1 joule per second (J/s).

Explanation:

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

[tex]KE = \frac{1}{2}mv^2[/tex]

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

[tex]U_s = \frac{1}{2}kx^2[/tex]

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

[tex]E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2[/tex]

We can simplify and solve for 'x'.

[tex]mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}[/tex]

Plug in the givens and solve.

[tex]x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}[/tex]

The net force on the crate will be 99 N.Option B is correct.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration.Mathematically in the different components and balancing the equation gets.Components in the x-direction.

When all the forces get resolved the y-direction of forces are;

300 sin 10° +N = 445

N = 445-295.4

N = 392.9 N

The normal force is 392.9 N.

The net force on the crate is found by resolving the force in the x-direction;

F = 300 cos 10° - μN

F= 295.44-196.45

F= 98.99

F=99

Hence, the net force on the crate will be 99 N.Option B is correct.

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Answer:

suna bro i am nepali and nepali are pro

The motion of the elevator can best be described as motion against gravity with negative acceleration.

Motion refers to a change in position of an object when it moves from one point another along a reference frame.

An object moving in a downward direction is moving in a positive direction and its velocity increases relative to the earth since gravity acts downwards.

An object moving in a upward direction is moving in a negative direction and its velocity decreases relative to the earth since gravity acts downwards.

Therefore, the motion of the elevator can best be described as motion against gravity with negative acceleration.

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Answer:

The last one - each  vector pointing towards the center of the circle must be the same length for uniform circular motion

The most electronegative atoms are located in the seventeenth group of the periodic table.

The term electronegativity refers to the ability of an atom to attract the electrons of a bond towards itself.  The more electronegative atom will have the electron pairs closer to itself.

Hence, in this case, the blue atom is more electronegative. The most electronegative atoms are located in the seventeenth group of the periodic table.

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The force exerted by student A with his scooter is 306 N and that of student B is 204 N.

The force exerted by each student is calculated from Newton's second law of motion.

F = ma

where;

F(A) = 127.5 x 2.4

F(A) = 306 N

F(B) = 120 x 1.7

F(B) = 204 N

Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.

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The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × [tex]\frac{0. 06}{0. 12* 1* 0. 9}[/tex]

friction factor = 0. 52 × [tex]\frac{0. 06}{0. 108}[/tex]

friction factor = 0. 52 × 0. 55

friction factor [tex]= 0. 289[/tex]

b. When V = 3mls

Friction factor = 0. 52 × [tex]\frac{0. 06}{0. 12 * 3* 0. 9}[/tex]

Friction factor = 0. 52 × [tex]\frac{0. 06}{0. 324}[/tex]

Friction factor = 0. 52 × 0. 185

Friction factor [tex]= 0.096[/tex]

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × [tex]\frac{1}{2*6. 6743 * 10^-11}[/tex] × [tex]\frac{1^2}{0. 120}[/tex] × [tex]\frac{1}{100}[/tex]

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = [tex]0. 096[/tex] × [tex]\frac{1}{1. 334 *10^-10}[/tex] × [tex]\frac{3^2}{0. 120}[/tex] × [tex]\frac{1}{100}[/tex]

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

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Answer:hight

Explanation:

Answer:

the anser is b

Explanation:

If the force applied on the object is in the direction of its motion, the speed of the object increases. If the force is applied in the direction opposite to the direction of motion, then it results in a decrease in the speed of the object.

The temperature reading in Fahrenheit scale when reading 310 K is  98.6F

Temperature is the degree of hotness or coldness of the object.

Given is a temperature in Kelvin scale = 310 K

Temperature on centigrade scale = 310 -273 = 37°C

Temperature conversion from Centigrade to Fahrenheit is

C/5 = F -32/9

Plug the value, we get

37/5 = F-32/9

66.6 = F -32

F = 32+66.6 =

F = 98.6

Thus, the temperature of 310 K in Fahrenheit is 98.6F.

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a. Ball P

Density is define as mass per unit volume or the quantity per unit volume, unit area, or unit length.

If [tex]m_{1}[/tex] is mass ,  [tex]V_{1}[/tex] is volume and [tex]D_{1}[/tex]  is density of ball P

[tex]m_{2}[/tex]  is mass ,  [tex]V_{2}[/tex] is volume and [tex]D_{2}[/tex] is density  of ball Q

It is given that  ,

[tex]V_{1}[/tex] = [tex]V_{2}[/tex];

But  [tex]m_{1}[/tex]= [tex]2m_{2}[/tex]

Based on above equation we have:

[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{2m_{2} }{V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 2[/tex]

Thus, density of ball P is twice that of Q.

Therefore ,

Ball  P will have a greater density than Q

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Answer:

[tex]28.0\; {\rm N}[/tex] to the right.

Explanation:

Since the sign is not moving, the net force on this sign should be [tex]0\; {\rm N}[/tex]. For that, the horizontal component ([tex]x[/tex]-component) of external forces on this sign should be [tex]0\; {\rm N}[/tex].

Sources of external forces on this sign include tension from the wires, as well as gravitational pull (weight) from the earth. The gravitational pull from the earth is entirely vertical ([tex]y[/tex]-component,) with a magnitude of [tex]0\; {\rm N}[/tex] in the horizontal direction. Thus, the only external forces on this sign in the [tex]x[/tex]-component would be from the two wires.

The question states that the [tex]x[/tex]-component of the force from the first wire is [tex]28.0\; {\rm N}[/tex] to the left. Thus, for the net force in the [tex]x[/tex]-direction to be [tex]0\; {\rm N}[/tex], the force from the other wire in the [tex]x\![/tex]-component needs to be [tex]28.0\; {\rm N}\![/tex] to the right (same magnitude but opposite direction.)

The I3 will be 158 A.

To calculate I3 firstly, V4 has to be calculated,

[tex]V_{4} =I_{4} R_{4}[/tex]

[tex]V_{4} = V_{2} / R_{4} + R_{5} * R_{4}[/tex]

[tex]V_{4} = 12 * 135 / 135+61[/tex]

[tex]V_{4} = 8.26V[/tex]

For I3,

[tex]I_{3} = R_{1} /(R1+R3 + (R1+R3)(R2+R6) * (V2 - V1 (R1+R2+R6/R1)[/tex]

[tex]I3=(61)/((61)(50)+(61+50)(141+141)) (12 -18 (1+(141+141)/61)) = -.158 A[/tex]

Hence, the current through I3 will be 158 A.

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The work done by the force to stretch the spring is determined as 12.5 J.

The work done by the force to stretch the spring is calculated as follows;

W = ¹/₂Fx

F = kx

W = ¹/₂F(F/k)

W = ¹/₂F²/k

W = ¹/₂ x (50²)/100

W = 12.5 J

Thus, the work done by the force to stretch the spring is determined as 12.5 J.

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Answer:

The force between two objects is calculated through the equation,

                       F = Gm₁m₂/d²

where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is the universal gravitation constant.

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,

                      2.5F = Gm₁m₂ / (D²)

                               D = 0.623d

Subsituting the known value of d,

                               D = 0.623(6.9 x 10^8) = 4.298 x 10^8 m

Answer:

di

Explanation:

The prefix is di. CO2 is carbon dioxide. di means 2.

Answer:

If theta is equal to 90, then both projectiles strike at the same distance from the projection point and projectiles are in air for the same time interval.

Explanation:

The force is reasonable for making fusion possible in the Sun is heat energy.

When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.

The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.

E = mc²

The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.

Thus, fusion requires heat energy to continue the reaction.

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The true statement about the accuracy and precision of measuring tools is Tools cannot be both accurate and precise. The correct option is C.

When in an experiment, a value is measured 5 times, then if the values measured are same for most of the time or like three times out of five, it said to be accurate. The phenomenon is accuracy.

Precision is about comparing the values to each other then find them near to each other.

Accuracy compares the experimental value to the theoretical value.

So, when all the values are close to each other but not nearest to the theoretical value, then it is said to be precise but not accurate.

Thus, the correct option is C.

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