Which one has more momentum if they are all traveling at the same speed?

1. A basketball
2. A golf ball
3. A tennis ball
4. A bowling ball​

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm   - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,the plank will not balance.

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. This moment tilts the plank in a clockwise direction, so the plank would tip up.

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

Answer:

Explanation:

Let the luminosity of the star be I and luminosity of the sun be Isun.

2.4 billion light years = 2.4 x 10⁹ light years .

brightness = luminosity / (distance)²

Given   Sun would have to be viewed from a distance of 1300 light-years to have the same apparent magnitude as 3C 273 so

For the sun

brightness = Isun / (1300 light years )²

For star

brightness = I / (2.4 x 10⁹ light years  )²

Both these brightness are same

Isun / (1300 light years )²  =  I / (2.4 x 10⁹ light years  )²

I = Isun x (2.4 x 10⁹ light years  )² / (1300 light years )²

= Isun x 3.4 x 10¹² .

Answer:

Approximately [tex]7.1 \times 10^{3}\; {\rm J}[/tex] (given: [tex]g = 9.8\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

To find the change in the gravitational potential energy ([tex]\text{GPE}[/tex]), use the formula:

[tex]\begin{aligned}& (\text{change in GPE}) \\ &= (\text{mass})\, (g)\, (\text{change in height})\end{aligned}[/tex].

Assume that gravitational field strength [tex]g[/tex] is constant (e.g., [tex]g = 9.8\; {\rm m\cdot s^{-2}}[/tex].) For an object of mass [tex]m[/tex], if the altitude of the object changes by [tex]\Delta h[/tex], the [tex]\text{GPE}[/tex] of that object would change by [tex]m\, g\, \Delta h[/tex].

In this question, the mass of Ben is [tex]m = 63.2\; {\rm kg}[/tex]. It is given that [tex]g = 9.8\; {\rm m\cdot s^{-2}} = 9.8\; {\rm N\cdot kg^{-1}}[/tex] and is constant. Since change in the altitude of Ben is [tex]\Delta h = 11.5\; {\rm m}[/tex], the change in the ([tex]\text{GPE}[/tex]) of Ben would be:

[tex]\begin{aligned} m\, g\, \Delta h &= (63.2\; {\rm kg}) \, (9.8\; {\rm N\cdot kg^{-1}})\, (11.5\; {\rm m}) \\ &\approx 7.1\times 10^{3}\; {\rm N\cdot m} = 7.1\times 10^{3}\; {\rm J} \end{aligned}[/tex].

Answer:

Average speed [tex]= 2.5[/tex] meter per second

Explanation:

Given

The total distance walked [tex]= 2250[/tex] meters

The total time taken to walk [tex]2250[/tex] meters is [tex]15[/tex] minutes

In one minute there are [tex]60[/tex] seconds

As we know Average speed is equal to total distance travelled divided by total time taken

Hence,

Average speed

[tex]\frac{2250}{15*60} \\2.5[/tex]meter per second

A unit vector must have length equal to 1. By calculating the length of (1, 1), we discover that (12 + 12) = 2 = 1.414, which is not the same as 1, is the length.

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The work done to the mower is the product of force and the distance covered by the object which is equal to 40,811 Joules.

Work done can be defined as the product of external force and the distance over which the force is being applied. Work is done on an object when a force is applied to an object and the object is moved through a particular distance.

From this we can see that:

X-Component:

cos(θ) = x/F

Then,

X= F cos(θ).…..(1)

Now putting the values of F and θ in equation (1) we get,

X = 72 N × cos(40)

X = 72 N × 0.766

X= 55.15 N this is the x-component force

The work done to the mower is:

W = F × d

W = 55.15 × 740

W = 40,811 Joules

The work done to the mower is 40811 Joules.

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Answer:

4 rad/s

Explanation:

The angular velocity is velocity over time:

[tex]\displaystyle{\omega = \dfrac{v}{r}}[/tex]

We know that the velocity is 20 m/s and the radius is half of diameter which is 10m/2 = 5 m. Hence, substitute in the formula:

[tex]\displaystyle{\omega = \dfrac{20 \: \text{m/s}}{5 \: \text{m}}}\\\\\displaystyle{\omega = 4 \: \text{rad/s}}[/tex]

Therefore, the angular velocity is 4 rad/s

The degree of data stability when the measurement is replicated under identical circumstances is known as reproducibility or reliability.

Reproducibility determines how an entire study an experiment can be replicated, whereas repeatability assesses the variation in data made by a single equipment or human under similar circumstances.

Science needs reproducibility because it enables more in-depth investigation, and replication validates our findings. There are several investigations and experiments, which result in a wide range of variables, unforeseen, and things that are either outside your influence or you cannot guarantee.

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Answer:

speed

Explanation:

you need to move fast so the tornado can't reach you

Answer:

19.284

Explanation:

The answer is right on Accellus.

Answer:

I believe it's applied force (push)

Answer:

The toy car

Explanation:

the real car is parked so yeah but maybe in some way technically the real car has more "momentum"

Explanation:

Dimensions help understand how physical amounts are related to their dependence on basic or fundamental volumes, i.e. how the body's dimensions rely on mass, time, length and temperature.

Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions.

Answer:

because they may not actually be moving be just percieved as so

Explanation:

Answer:

i think that maybe there arent moving but there on something thats moving

Explanation:

Answer: 19.80 cm

Explanation:

Given

focal length [tex]f=6.14\ cm[/tex]   (as focal length is positive, it is converging lens)

Image distance [tex]v=1.5f[/tex]

[tex]v=1.5\times 6.14\\v=9.21\ cm[/tex]

using lens formula

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

insert values

[tex]\dfrac{1}{u}=\dfrac{1}{6.14}-\dfrac{1}{8.9}\\\\\dfrac{1}{u}=0.1628-0.1123\\\\\dfrac{1}{u}=0.0505\\\\u=19.80\ cm[/tex]

Thus, the distance of the object is 19.80 cm

Given:

Now,

The image distance will be:

→ [tex]v = 1.5 f[/tex]

     [tex]= 1.5\times 6.14[/tex]

     [tex]= 9.21 \ cm[/tex]

By using the lens formula, we get

→ [tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]

By putting the values, we get

→ [tex]\frac{1}{u} = \frac{1}{6.14} - \frac{1}{8.9}[/tex]

   [tex]\frac{1}{u} = 0.1628-0.1123[/tex]

   [tex]\frac{1}{u} = 0.0505[/tex]

   [tex]u = 19.80 \ cm[/tex]

Thus the answer above is correct.    

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The velocity of the experimental vehicle after 2 seconds is 14 m/s.

The displacement of the experimental vehicle after 2 seconds is 14 m.

The velocity of the experimental vehicle is the speed of the in a given direction starting from rest.

The velocity, v, of the experimental vehicle after 2 seconds is calculated using the formula below:

v = u + at

where;

u is the initial velocity = 0 m/s

a is the acceleration = 7 m/s²

t is time = 2 seconds

v = 0 + 7 * 2

v = 14 m/s

The displacement, s, of the vehicle is calculated below as follows:

v² = u² + 2as

14² = 0² + 2 * 7 * s

s = 196 / 14

s = 14 m

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Answer:

S = 2 * pi * 1 m = 6.28 m = distance traveled

V = S / T   or   T = S / V = 6.28 m / 5 m/s = 1.26 sec

This will be the time for 1 revolution or the period of the motion.    

Answer:

The moon is 1,079.4 mi.

Mars is 2,106.1 mi

Multiply your weight by the moon's gravity relative to earth's, which is 0.165. Solve the equation. In the example, you would obtain the product 22.28 lbs. So a person weighing 135 pounds on Earth would weigh just over 22 pounds on the moon

Being that Mars has a gravitational force of 3.711m/s2, we multiply the object's mass by this quanitity to calculate an object's weight on mars. So an object or person on Mars would weigh 37.83% its weight on earth.

Explanation:

~Hope this helps

Answer:

The toy car. An object that isn't moving has no momentum

Explanation:

Based on the data given:

The change in momentum of the object equals the impulse experienced by the object

Mathematically,

Impulse = change in momentum

Impulse = F • t

change in momentum = m • Δ v

F • t = m • Δ v

where

F is the force

t is the time the force acts

m is the mass of the object

Δv is the change in velocity

For the given values:

change in momentum of the object = 7.36 * 24.7

change in momentum = 181.8 N.s

Change in velocity, Δv = change in momentum / mass

Change in velocity, Δv =  181.8 / 2.1

Change in velocity, Δv = 86.6 m/s

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Answer:

AS

Explanation:

Answer:

Explanation:

An electron accelerates through a 12.5 V potential difference, starting from rest, so it will acquire kinetic energy of 12.5 eV .

In hydrogen atom energy of n th orbit in terms of eV is given as follows

En = -13.6 / n² eV

Total energy of 1 st orbit  E₁ = - 13.6 eV

Total energy of 2 st orbit E₂ = - 13.6 eV / 2² = - 3.4 eV

Total energy of 3 st orbit E₃ = - 13.6 eV / 3² = - 1.5  eV

Total energy of 4 st orbit E₄ = - 13.6 eV / 4² = - 0.85 eV

E₄ - E₁ = 13.6 - 0.85 = 12.75 eV

E₃ - E₁ = 13.6 - 1.5 = 12.10 eV

E₂ - E₁ = 13.6 - 3.4 = 10.2 eV .

The electron has energy of 12,5 eV so it can excite electron from E₁ to E₃ . .

Jump possible = E₃ to E₂ , E₂ to E₁ and E₃ to E₁

Energy of E₃ to E₂ = 3.4 - 1.5 eV = 1.9 eV

wavelength = 1237 / 1.9 nm = 651 nm

E₃ - E₁ = 13.6 - 1.5 = 12.10 eV

wavelength  = 1237 / 12.10  nm = 102.23 nm

E₂ - E₁ = 13.6 - 3.4 = 10.2 eV

wavelength  = 1237 / 10.2  nm = 121.27  nm

wavelength of photon possible are 651 nm , 121.27  nm , 102.23 nm .

(a) The kinetic energy of the ball just before it hits the ground is 78.4 J.

(b) The kinetic energy of the ball at a height of 1.2 m is 31.36 J.

The kinetic energy of the ball just before it hits the ground is calculated as follows;

K.E ( bottom) = P.E (top)

P.E (top) = mgh

where;

P.E (top) = 4 x 9.8 x 2

P.E (top) = 78.4 J

The change in the potential energy of the ball during its motion is calculated as follows;

ΔU = P.Ef - P.E i

where;

ΔU = 0 J - 78.4 J

ΔU = -78.4 J

The kinetic energy of the ball at a height of 1.2 m is the change in the potential energy of the ball and it is calculated as follows;

K.E = -ΔU

K.E = -mg (hf - hi)

K.E = (4 x 9.8)(2m - 1.2 m)

K.E = 31.36 J

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Answer:

near a flame and a hot plate

Explanation:

Answer:

newtons

Explanation:

In the metric system of units, where force is measured in newtons (abbreviated N), work is measured in newton-meters (N-m). For reference, a newton is roughly equal to the force exerted on your hand by a baseball.

Answer:

newtons

In the metric system of units, where force is measured in newtons (abbreviated N), work is measured in newton-meters (N-m). For reference, a newton is roughly equal to the force exerted on your hand by a baseball.

Answer:

A) It is the length from the midpoint to the crest.

Explanation:

The correct statement about the amplitude of any transverse wave is: It is the length from the midpoint to the crest. The correct option is A.

A transverse wave is a type of wave in which the displacement of the medium is perpendicular to the direction of wave propagation.

Examples of transverse waves include light waves and water waves in which the surface of the water oscillates up and down while the wave moves horizontally.

The properties of a transverse wave include:

Amplitude: The maximum displacement of the medium from its rest position. This represents the intensity or strength of the wave.

Wavelength: The distance between two adjacent points in the wave that are in phase with each other, for example, between two consecutive crests or troughs. It is typically represented by the symbol λ.

Frequency: The number of complete wave cycles that pass a point in a given amount of time, usually measured in hertz (Hz). The frequency is inversely proportional to the wavelength and is represented by the symbol f.

Period: The time taken for one complete wave cycle to pass a given point, usually represented by the symbol T. The period is directly proportional to the wavelength and inversely proportional to the frequency.

Speed: The speed at which the wave propagates through the medium, usually represented by the symbol v. The speed is directly proportional to the frequency and wavelength.

These properties are related to each other by the wave equation:

v = fλ,

Where v = the speed of the wave,

f = the frequency,

and λ = the wavelength.

Therefore, The correct statement about the amplitude of any transverse wave is A. It is the length from the midpoint to the crest.

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Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, [tex]\lambda=284\ nm[/tex]

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

[tex]n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm[/tex]

So, the wavelength in vacuum is equal to 428.8 nm.