a convex lens has a focal length of 0.5 it is combined with a second lens so combination has power 2.5 diopter which of the following could be second lens

A) concave lens focal length 2m
B) convex lens focal length 0.5m
C) concave lens 0.5m
D) convex lens 2m

Answer:

d and c are the answers :)

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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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

[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]

Answer:

Nuclear energy is the energy stored in atoms that can produce electricity.

Hope that helps. x

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.

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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: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.

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Answer:

1kW = 1000W

600W = 0.6kW

Cost of electric bill = 0.6kWh × 24 × 30 × $0.05

                               = $21.60

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

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.

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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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

[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]

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:

The amount of heat generated in the system is calculated as follows;

Q = mcΔT

where;

If 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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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.

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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Answer:

So correct sentence is:

So correct sentence is:

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

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:

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

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.

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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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.

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

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.)

Answer: 25N to the right

Explanation: 35N - 10N = 25N

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

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.

Answer:

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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:

[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

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