Which statement about allele combinations is accurate?

Question 5 options:

Offspring get two alleles for each trait – one from each parent.


If the dominant allele is passed from either parent, the offspring will show the dominate trait.


If each parent gives a different allele, the genotype is referred to as heterozygous.


These statements are all true.

The acceleration of a 50 g object pushed with a force of 0.5 N is 10 m/s².

To find the acceleration of the object, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:

F = m * a

Given:

Force (F) = 0.5 N

Mass (m) = 50 g = 0.05 kg

Substituting the given values into the equation, we have:

0.5 N = 0.05 kg * a

To find the acceleration (a), we rearrange the equation:

a = F / m

a = 0.5 N / 0.05 kg

a = 10 N/kg

Since acceleration is measured in meters per second squared (m/s²), we convert the unit of N/kg to m/s²:

1 N/kg = 1 m/s²

Therefore, the acceleration of the 50 g object pushed with a force of 0.5 N is 10 m/s².

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If the Amplitude of the wave is doubled the Energy of the wave is quadrupled.

For a wave with Energy \(E\) and Amplitude \(A\) :

\(E \propto A^2\\ \Rightarrow E=kA^2\qquad(1)\)

where \(k\) is the proportionality Constant

If Amplitude is doubled, Then the Energy \(E'\) is given as:

\(E'=k(2A)^2 \\ \Rightarrow E'=4(kA^2) \qquad (2)\)

Using equation (1) in (2), we get:

\(\boxed{E'=4E}\)

So, If the Amplitude of the wave is doubled the Energy of the wave is quadrupled.

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It's important to know that when poles have the same nature, the force is repulsive. But, when the poles have different nature, the force will be attractive.

Answer: A literature review consists of an overview, a summary, and an evaluation (“critique”) of the current state of knowledge about a specific area of research.

Explanation:

Answer:

13 hz

Explanation:

Speed of sound = 1522

Dolphin swims off at 8.0 = Cs

Frequency of clicks = Fo = 2500

Fo = fs(c/c+cs)

2500 = fs(1522/1522+8)

2500 = fs(1522/1530)

2500 = 0.99477fs

Fs = 2500/0.99477

= 2513.1

From here, we find the difference between this frequency and the number of clicks per second

= 2513.1 - 2500

= 13.1 hz

Approximately 13 Hz

Answer: what’s the answer

Explanation:

Answer:1.9 atm

Explanation: I stole it from someone O-O

So it could be wrong

Answer:

B. Containing charged regions

Explanation:

The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions

So in the given case, the option b would be contributed to the molecules that have intermolecular forces

hence, the option b is correct

The resultant velocity of the boat is 7.5 km/h.

The resultant displacement of the boat is calculated as follows;

Sum of the vertical displacement of the boat is calculated as;

∑Fy = -24 km sin(50)  + 8 km sin(60)

∑Fy = -11.5 km

Sum of the horizontal displacement of the boat is calculated as;

∑Fx = -24 km cos(50)  - 8 km cos(60)

∑Fx = -19.4 km

The resultant displacement is calculated as follows;

d = √ (-11.5² + 19.4²)

d = 22.55 km

The resultant velocity of the boat is calculated as follows;

v = ( 22.55 km ) / ( 3 hrs )

v = 7.5 km/h

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Answer:

Lightning begins as static charges in a rain cloud.Water droplets in the bottom part of the cloud are caught in the updrafts and lifted to great heights where the much colder atmosphere freezes them. Meanwhile, downdrafts in the cloud push ice and hail down from the top of the cloud.

Explanation:

Answer:

when it rains the clouds get darker as they get wet because its stormy which causes lightning to come in trying to make the clouds start getting darker.

Explanation:

The energy is lost in friction between the box and the inclined plane while pushing the box up the plane and into her truck.

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Based on the mass and luminosity of the Sun, the time required to convert of its original hydrogen into helium is 7.505 × 10^10 years.

Nuclear fusion is the process whereby nucleus of lighter atoms are fused together to produce heavier atoms with the release of large amounts of energy.

Nuclear fusion occurs at the core of the sun where four hydrogen atoms are fused to form one helium atom with the release considerable energy, and this is known as hydrogen fusion.

The mass of the Sun = 2.0 × 10^30 kg

The Sun’s hydrogen mass is 0.71 × 2.0 × 10^30 kg = 1.42 × 10^30 kg.

The Sun converts 6 × 10^11 kg of hydrogen into helium per second.

Time required to convert all of the Sun's hydrogen to helium = 1.42 × 10^30 kg/6 × 10^11 kg/s

Time required = 2.367 × 10^18 seconds

Converting to years:

3.154 × 10^7 seconds = 1 year

2.367 × 10^18 seconds = 2.367 × 10^18 seconds * 1 year/3.154 × 10^7 seconds

Time required = 7.505 × 10^8 years

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The magnitude of the vector is approximately 35.85 m.

The angle that this vector makes with the positive x-axis is approximately 32 degrees.

To find the magnitude of the vector with components Ax=29 m and Ay=18 m, we use the Pythagorean theorem:

|A| = √(Ax^2 + Ay^2)

|A| = √(29^2 + 18^2)

|A| = √(841 + 324)

|A| = √1165

|A| =  34.13 m

To find the angle that this vector makes with the positive x-axis, we can use the inverse tangent function:

θ = tan^-1(Ay/Ax)

θ = tan^-1(18/29)

θ = 31.82 degrees

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Heat transfer by radiation and convection causes the surface water of the pool to be warm which decreases with depth.

There are two heat transfer process responsible for variation in temperature of water at different depth of a pool, they include;

The surface of the water absorbs heat from the air and the sun. The heat from the sun is transferred to the pool through radiation.

The heat from the air is transferred to the pool through convection.

Thus, these two heat transfer processes  causes the surface water of the pool to be warm which decreases with depth.

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Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = \(F_{average}\) × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

\(F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}\)

∴ \(F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N\)  

The average force on the right sled applied by the cat while landing, \(\mathbf{F_{average}}\) = 922.625 N

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The answers are 4. The distance from where the ball was kicked is 38.06 meters, 5. The speed of the ball 2.5 seconds after it was kicked is 13.82 m/s, and  6. The ball is 21.88 meters above the ground 2.5 seconds after it is kicked.

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:  If you are traveling at a speed of 60mph, you will go 220 feet.

Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.

Answer:

75m/s

Explanation:

...................

We can calculate the time taken by the compass to hit the ground by using kinematic equations of motion. The motion of the compass is a free-fall motion since it is only under the influence of gravity. When the compass is dropped, it is initially at rest.

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The kinetic energy in terms of the amplitude of oscillation is,

where k is the spring constant and x is the position,

The potential energy in terms of the amplitude of oscillation is,

The value of position when the kinetic energy of the oscillation is 3 times the potential energy is,

By simplifying,

Thus, the position at which the kinetic energy becomes 3 times the potential energy is half of the amplitude.

Answer:

No kinetic energy-because there is no movement

Explanation:

It states that the monk is at complete rest so there is no movement.

Explanation:

To determine the effort needed to lift a load of 200N, given a velocity ratio of 5 and an efficiency of 80%, we can use the formula:

Efficiency = (Output Work / Input Work) * 100

Efficiency can also be calculated as the ratio of the output force to the input force. In this case, the output force is the load being lifted (200N), and the input force is the effort required.

Given that the velocity ratio is 5, it means that for every 5 units of distance the effort moves, the load moves 1 unit of distance. This implies that the effort is exerted over a greater distance than the load.

Let's denote the effort force as "E" and the distance moved by the effort as "dE." Similarly, the load force is "L," and the distance moved by the load is "dL."

Using the velocity ratio, we have the following relationship:

dE / dL = 5

Now, we can calculate the input work (Wi) and the output work (Wo):

Input Work (Wi) = Effort (E) * Distance moved by the effort (dE)

Output Work (Wo) = Load (L) * Distance moved by the load (dL)

Given that the efficiency is 80%, we can rewrite the formula for efficiency as:

0.80 = (Wo / Wi) * 100

Now, let's solve for the effort (E) using the given values:

Load (L) = 200N

Efficiency = 0.80

Velocity Ratio = 5

First, calculate the output work (Wo):

Wo = Load (L) * Distance moved by the load (dL)

Since the velocity ratio is 5, the distance moved by the load (dL) will be 1/5 of the distance moved by the effort (dE):

dL = (1/5) * dE

Wo = L * (1/5) * dE

Wo = 200N * (1/5) * dE

Wo = 40N * dE

Next, calculate the input work (Wi):

Wi = Effort (E) * Distance moved by the effort (dE)

Wi = E * dE

Now, substitute the values into the efficiency formula:

0.80 = (Wo / Wi) * 100

0.80 = (40N * dE) / (E * dE) * 100

0.80 = 40 / E * 100

0.80 * E = 40

E = 40 / 0.80

E = 50N

Therefore, the effort needed to lift a load of 200N with a velocity ratio of 5 and an efficiency of 80% is 50N.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

16 reflection,incidence

Answer:

B.

Divide the height of the valley walls by the rate of erosion.

Explanation:

There is a relationship between the rate of erosion and the hieght at which it is eroded according to Newton's law of motion. In the case of the scenario above, the best way to determine the time the river started carving the valley would be the division of the height of the valley walls by the rate of erosion.

Answer:

"12.122 m/s²" is the appropriate solution.

Explanation:

The given values in the question are:

Length,

l = 49.0 cm

or,

 = \(49.0\times 10^{-2} \ m\)

Time taken,

\(T = \frac{125}{99.0}\)

   \(=1.2626 \ s\)

Now,

As we know,

⇒  \(T = 2 \pi\sqrt{\frac{l}{g} }\)

or,

⇒  \(T^2=4 \pi^2[\frac{l}{g} ]\)

     \(g = 4\pi^2[\frac{l}{T^2} ]\)

By substituting the above given values, we get

        \(=\frac{4\times (3.14)^2\times (49.0\times 10^{-2})}{(1.2626)^2}\)

        \(=\frac{39.4384\times 49.0\times 10^{-2}}{1.59415876}\)

        \(=\frac{19.324816}{1.59415876}\)

        \(=12.122 \ m/s^2\)

Answer:

speed=30.41, height=94.27

Explanation:

since the motion is vertical upward means the motion is against gravitational force and the initial velocity of the ball was zero.

I see six (6) loops.

I attached a drawing to show where I get six loops from.