A car accelerates from rest at a constant rate of 2 m/s^2 for 5 s. what is the speed of the car at the end of that time? g

Answer:

(2) 2s

Explanation:

Remember that the time that it takes an object to fall from a certain distance is only determined by the force with which the object is pulled towards the center of the earth which is gravity, so any object with 0 velocity will drop at the same rate to the ground when dismissing resistance from the air, in to calculate this you just have to use the next formula:

[tex]H=\frac{g*t^2}{2}\\ t=\sqrt{\frac{2H}{g} }[/tex]

So we just insert the data that we have into the formula:

[tex]t=\sqrt{\frac{2H}{g} }\\t=\sqrt{\frac{2*19,6}{9,81} }\\t=\sqrt{4}\\ t=2 seconds[/tex]

Final answer:

The distance from the cannon to the ship is 32.0 km, but to calculate initial velocity and maximum height of the shell, additional information such as angle of launch or time is required. The Earth's curvature slightly affects the height of the ocean's surface relative to the ship over long distances.

Explanation:

Assuming the dread pirate Roberts never misses, the distance from the end of the cannon to the ship trying to be hit is the maximum distance the cannon shell can travel, which is 32.0 km when neglecting the dimensions of the cannon and air resistance. To calculate the initial velocity of the shell, we would use the kinematic equations for projectile motion. However, the question does not provide angle of launch or time required for the calculation, so further information would be needed to complete part (a).

For part (b), the maximum height reached by the shell cannot be calculated without additional information such as the launch angle or time of flight. For part (c), the curvature of the Earth impacts the level of the ocean's surface in relation to a straight line extending from the ship. Using the Earth's radius (6.37×10³ km), we can apply the principles of geometry to find the drop in height over a distance of 32.0 km, with the assumption that the ocean surface follows the curvature of the Earth.

The distance of every object in a round trip is always positive

Distance is a scalar quantity, the total distance of an object in any trip is the total path covered by the object from the starting point to the finish point.

On the other hand, the displacement of the object will be zero. Displacement is the change in the position of an object.

Thus, we can conclude that the distance of every object in a round trip is always positive while the displacement is zero.

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We know that in an elastic collision the moments of the both the objects is conserved i.e

Intial momentum of object1 + Intial momentum of object2 =Final momentum of object1 + Final momentum of object2

Intial momentum of object1 =12 kg.m/s

Intial momentum of object2 = P (assume that P is the momentum)

Final momentum of object1= 14 kg.m/s

Final momentum of object2 =16 kg.m/s

On substituting all values we get

12 + P =14 kg.m/s + 16 kg.m/s

P = -18 kg.m/s

(-ve sign indicates the second object was moving in the opposite direction to the object1 before collision )

Therefore the initial momentum of the second object was 18 kg.m/s.

i just did it and the answer us A

The greatest biodiversity on earth is found in the tropical rainforest biome.

Biome is actually another name for ecosystem. Rainforests are basically the wettest ecosystems and very diverse due to some reasons such as very high annual rainfall, high average temperatures, nutrient-poor soil, and high levels of biodiversity. Biomes or ecosystems are characterized by their climate and on that basis we can find which type of animals and plants can be found there. The greatest biodiversity on earth is found in the tropical rainforest biome.

The greatest biodiversity on earth is found in the tropical rainforest biome. Hence, option D is correct.

Biodiversity or biological diversity is the measure of variation at the species, genetic, and ecosystem levels. It comprises all the different kinds of life and supports life on Earth.

Biodiversity is important because it supports the entire life on the Earth including plants, animals, microorganisms, etc. Without biodiversity, a healthy ecosystem is not possible. Biome is called an ecosystem. Rainforest has the wettest ecosystems and has a higher annual rainfall and nutrient-rich soil.

The greatest biodiversity on earth is found in the tropical rainforests biome and hence, the ideal solution is option D.

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Answer:

D.  1.4 × 10^2 kilogram meters/second

Answer: 3 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:

[tex]p_i = p_f[/tex]

The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore

[tex]p_i = m_c u_c = (0.5 kg)(3 m/s)=1.5 kg m/s[/tex]

So, the final total momentum will also be

[tex]p_f = 1.5 kg m/s[/tex]

And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:

[tex]p_f = m_s v_s[/tex]

from which we find the velocity of the solid ball

[tex]v_s = \frac{p_f}{m_s}=\frac{1.5 kg m/s}{0.5 kg}=3 m/s[/tex]

3, just did the assignment

The mass (kg) of benzene is about 4470 kg

This problem is about Density.

Density is the ratio of mass to the volume of the object.

[tex]\large {\boxed {\rho = \frac{ m }{ V } } }[/tex]

ρ = density of object ( kg / m³ )

m = mass of object ( kg )

V = volume of object ( m³ )

Given:

Diameter of Cylinder = d = 2.00 m

Radius of Cylinder = R = d/2 = 2.00/2 = 1.00 m

Length of Cylinder = L = 4.00 m

Liquid Depth = H = 0.85 m

Density of Benzene = ρ = 0.879 g/cm³ = 879 kg/m³

Unknown:

mass of benzene  = m = ?

Solution:

This problem is about Liquid Volume in Partially Filled Horizontal Tanks

Firstly we will calculate the volume of Benzene by using following formula:

[tex]V = A \times L[/tex]

[tex]V = ( \texttt{Area of Sector - Area of Triangle} ) \times L[/tex]

[tex]V = [ R^2 \cos^{-1}(\frac{R - H}{R}) - (R - H)\sqrt {(2RH - H^2)} ] L[/tex]

[tex]V = [ 1^2 \cos^{-1}(\frac{1 - 0.85}{1}) - (1 - 0.85)\sqrt {(2(1)(0.85) - 0.85^2)} ] 4[/tex]

[tex]V = [ \cos^{-1} (0.15) - 0.15 \sqrt{ 0.9775} ] 4[/tex]

[tex]V \approx \boxed {5.0877 ~ m^3}[/tex]

[tex]m = \rho \times V[/tex]

[tex]m = 879 \times 5.0877[/tex]

[tex]m \approx \boxed {4470 ~ kg}[/tex]

Grade: High School

Subject: Mathematics

Chapter: Density

Keywords: Temperature , Density , Iron , Sphere , Volume , Mass

Scientists have proven that genes play no role in self-esteem is a FALSE statement. In fact, most mental health (including self-esteem) has many connections with genetics.

The OXTR gene is said to be related to our self-esteem. Just a fun fact:

Those with 1 or 2 copies of the OXTR with an "A" allele gene are shown to be the more "negative" type of people with lower self-esteems. Those with 2 copies of the OXTR with a "G" allele gene were said to be more optimistic.

The statement that genes do not affect self-esteem is false. Research has found a connection between certain genes and levels of self-esteem, although genes are not the only influencing factors.

The statement that genes play no role in self-esteem is False. It has been scientifically proven that genes play a role in determining an individual's level of self-esteem. Through research studies, it has been observed that there is a connection between specific genes and self-esteem. Such studies involve analyzing the DNA of different individuals and examining the likelihood of these individuals having higher or lower self-esteem, depending on the presence of certain genes.

Research suggests that genes can have an influence on certain personality traits which are associated with self-esteem. For instance, individuals who are naturally more outgoing or socially inclined due to their genetic makeup might have a higher self-esteem than those who are introverted. However, it's important to note that genes are not the only factors that influence self-esteem - environmental factors also play a significant role.

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Answer: The velocity of the object will be 8m/s.

Explanation:

Force exerted on the object = 80 N

Distance displaced by object by the action of force = 4 m

Mass of the object =  10 kg

Velocity gained by the object = v

Kinetic energy of the object =  Work done on the object

[tex]\frac{1}{2}mv^2=Force\times displacement[/tex]

[tex]\frac{1}{2}\times 10\times v^2=80 N\times 4 m[/tex]

[tex]v^2=64 m^2/s^2[/tex]

[tex]v=8 m/s[/tex]

The velocity of the object will be 8m/s.

Final answer:

The number 6.3 is expressed in scientific notation as 6.3×10^0.

Explanation:

To express the number 6.3 in scientific notation, we need to follow a standard format where the number is between 1 and 10 followed by an exponent of 10.

For 6.3, this is already the case, so it can be expressed as 6.3×10^0 because any number raised to the power of 0 is equal to 1, so this does not change the value of 6.3.

Therefore, when writing in scientific notation, we are just acknowledging that we could move the decimal zero places and maintain the same value.

The examination of color and streak are the physical tests that are performed on the minerals to know their exact origin.

Hence, this can be stated that a streak is useful to give true color of the mineral.

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The energy from stars is released through a process called nuclear fusion, where hydrogen atoms combine to form helium and release a tremendous amount of energy. This energy is in the form of electromagnetic radiation, including visible light, infrared radiation, ultraviolet radiation, and X-rays. Stars release this energy continuously throughout their lifetimes.

The energy from stars is released through a process called nuclear fusion. In the core of a star, hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. This energy is in the form of electromagnetic radiation, which includes visible light, infrared radiation, ultraviolet radiation, and X-rays.

During nuclear fusion, the mass of the combined helium nucleus is slightly less than the mass of the original hydrogen atoms. This mass difference is converted into energy according to Einstein's famous equation, E=mc^2, where E represents energy, m represents mass, and c represents the speed of light.

Stars release this energy continuously throughout their lifetimes, providing the heat and light that sustains life on Earth and allowing astronomers to study distant objects in the universe.

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First let us convert the distance into meters.

distance = 1600 km = 1,600,000 m

Then we get the maximum time by dividing the distance with the smallest movement rate possible, that is:

maximum time = 1,600,000 m / (50 m / year)

maximum time = 32,000 years

Final answer:

Using the minimum movement rate of 50 meters per year, the maximum amount of time required for an ice sheet to travel 1600 kilometers from Hudson Bay to Lake Erie is calculated to be 32,000 years.

Explanation:

The question revolves around calculating the maximum amount of time required for an ice sheet to move from the southern end of Hudson Bay to the south shore of present-day Lake Erie, a distance of 1600 kilometers, given varying movement rates.

First, convert kilometers to meters since the movement rates are given in meters per year. The distance is 1,600 kilometers, which equals 1,600,000 meters. Since we are interested in the maximum amount of time required, we will use the minimum speed of 50 meters per year. This will yield the longest possible time it would take for the ice sheet to cover this distance.

To find the time, we use the formula: Time = Distance / Speed. Substituting the given values: Time = 1,600,000 meters / 50 meters per year = 32,000 years.

Therefore, at a movement rate of 50 meters per year, the maximum amount of time required for an ice sheet to travel from Hudson Bay to Lake Erie would be 32,000 years.

We can use the temperature coefficient of resistance to determine the resistance of the nichrome wire at 0.0 °C. The temperature coefficient of resistance (α) is the amount by which the resistance of a material changes per degree Celsius of temperature change.

Given information:

Resistance of the nichrome wire at 11.5°C = 200.00 Ω

Temperature at which resistance is to be found = 0.0°C

We can use the following formula to find the resistance of the nichrome wire at 0.0°C:

R₂ = R₁ [1 + α (T₂ - T₁)]

Where,

R₁ = Resistance of the wire at temperature T₁

R₂ = Resistance of the wire at temperature T₂

α = Temperature coefficient of resistance

T₁ = Temperature at which R₁ is given

T₂ = Temperature at which R₂ is to be found

Since we are given the resistance of the nichrome wire at 11.5°C, we can take this as R₁ and T₁ as 11.5°C. We also know that the temperature coefficient of resistance for nichrome wire is 0.0004 per °C.

Substituting the given values into the formula, we get:

R₂ = 200.00 Ω [1 + (0.0004/°C) (0.0°C - 11.5°C)]

R₂ = 200.00 Ω [1 - 0.0046]

R₂ = 200.00 Ω (0.9954)

R₂ = 199.08 Ω

Therefore, the resistance of the nichrome wire at 0.0°C is 199.08 Ω.

Nichrome wire resistance at 0.0°C is 199.08 Ω.

Calculate nichrome wire resistance at 0.0°C using its temperature coefficient and resistance at 11.5°C.

The resistance R of a conductor at a temperature T is given by the formula:

[tex]\[ R_T = R_0 \left(1 + \alpha \Delta T\right) \][/tex]

Rearrange the formula to find [tex]\( R_0 \)[/tex] (resistance at 0.0°C):

[tex]\[ R_0 = \frac{R_{11.5}}{1 + \alpha \Delta T} \][/tex]

[tex]\( \Delta T \)[/tex]= 11.5 °C - 0.0°C = 11.5°C

Substitute the values:

[tex]\[ R_0[/tex]=  200.00Ω/(1 + (0.0004°C* 11.5°C))

[tex]\[ R_0[/tex] = 200.00Ω/(1 + 0.0046)

[tex]\[ R_0[/tex] = 200.00Ω/(1.0046)

[tex]\[ R_0[/tex] = 199.08 Ω

It is the second one or the one with the letter F

Answer:

It takes 2.04s to get to the maximum height

Explanation:

This is a vertical throw problem so it can be  treated as a uniform accelerated rectilinear motion. For computing time we are going to use the formula:

[tex]v_{f}=v_{o}+g*t[/tex]

where[tex]v_{f}[/tex] is the final velocity, [tex]v_{o}[/tex] is the initial velocity, [tex]t[/tex] is the time and, [tex]g[/tex] is the gravity.

For solving this kind of problems we need at least three values. The values we have are:

Now:

[tex]v_{f}=v_{o}+g*t[/tex]

[tex]v_{f}-v_{o}=g*t[/tex]

[tex]\dfrac{v_{f}-v_{o}}{g}=t[/tex]

[tex]\dfrac{0-20}{-9.8}=t[/tex]

[tex]t=2.04s[/tex]

The two football players move together at a velocity of 5.99 m/s immediately after the collision. This value was obtained by conserving the momentum of the system: calculating individual momenta before the collision, summing them to obtain total momentum, and then dividing this by the total mass of both players. This is a typical conservation of momentum problem in high school level physics.

The situation you're describing is a perfect example of a conservation of momentum problem in physics. In this problem, the two football players can be seen as a system, and their individual momenta before the tackle add up to yield the total momentum of the system after the tackle, when they're moving together.

Momentum is calculated as mass times velocity, so we first calculate the momentum of each player before the collision. For the linebacker: 120kg× 8.6m/s = 1032 kg×m/s north. For the halfback: 75kg× 7.4m/s = 555 kg×m/s east.

These two momentum vectors form a right triangle, with the hypotenuse representing the  result vector or total momentum of the system after the tackle. We can use the Pythagorean theorem to calculate the magnitude of this vector: sqrt((1032²)+(555²)) = 1169 kg*m/s.

Since the players are moving together after the collision, the mass we use to find the final velocity should be the total mass of both players: 120kg + 75kg = 195 kg. The magnitude of the velocity at which the two players move together immediately after the collision can then be obtained by dividing the total momentum by the total mass: 1169kg×m/s / 195kg = 5.99 m/s.

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The steel guitar string stretches by 15 mm under a tension of 1500 N.

To determine how far a steel guitar string stretches under a tension force, we use principles from materials science, specifically Young's modulus and Hooke's law.

Identify the given values:

Initial length, l = 1.00 m

Cross-sectional area, a = 0.500 mm² = 0.500 × 10⁻⁶ m²

Young's modulus, Y = 2.0 × 1011 N/m²

Tension force, F = 1500 N

Use the formula for elongation:

δl = (F × l) / (Y × a)

Substitute the given values into the formula:

[tex]\delta l = \frac{1500 \, \text{N} \times 1.00 \, \text{m}}{2.0 \times 10^{11} \, \text{N/m}^2 \times 0.500 \times 10^{-6} \, \text{m}^2} \\[/tex]

Calculate the result:

[tex]\delta l = \frac{1500 \times 1.00}{2.0 \times 10^{11} \times 0.500 \times 10^{-6}} \\[/tex]

[tex]\delta l = \frac{1500}{1.0 \times 10^5} = 0.015 \, \text{m} = 15 \, \text{mm}[/tex]

The string stretches by 15 mm when a tension of 1500 N is applied.