Suppose that you run along three different paths from location A tolocation B. Along which path(s) would your distance traveled be different than your displacement
(Question 2) please help

Answers

Answer 1

Answer:

Answer:

Path 3 and Path 1

Explanation:

Along Path1andPath3, the distance travelled will be different than the displacement.

In Path1 and Path3 the distance travelled will be more than the displacement. Whereas, in Path2, the displacement will be as same as the distance travelled because in path 2, the distance travelled itself is the shortest distance from initial point ( A ) to final point ( B ).

But, inPath1 and Path3, the total distance travelled isn't the shortest distance from initial point ( A ) to final point ( B ), hene displacement and distance travelled will be different.

$\rule{200}2$

Answer 2

Answer:

Along Path 1 and path 3 your distance traveled be different than your displacement.

Distance is the total length of the path traveled by an object. It is a scalar quantity, meaning it has only magnitude and no direction.

Displacement is the change in the object's position from its initial to its final position. It is a vector quantity, meaning it has both magnitude and direction.

For example, if an object moves in a circle, the distance traveled will be equal to the circumference of the circle, but the displacement will be zero, since the object returns to its initial position.

Another example is if an object moves back and forth along a straight line, the distance traveled will be twice the length of the line, but the displacement will be zero, since the object returns to its initial position.

In general, the distance traveled will always be greater than or equal to the displacement. This is because the distance traveled includes all of the path that the object takes, while the displacement only includes the change in the object's position.

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What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?

Answers

Answer:

Explained

Explanation:

In order to retain atmosphere a planet needs to have gravity. A gravity sufficient enough to create a dense atmosphere around it, so that it can retain heat coming from sun. Mars has shallow atmosphere as its gravity is only 40% of the Earth's gravity. Venus is somewhat similar to Earth but due to green house effect its temperature is very high. Atmosphere has a huge impact on the planets ability to sustain life. Presence of certain kind gases make the atmosphere poisnous for life. The atmosphere should be such that it allows water to remain in liquid form and maintain an optimum temperature suitable for life.

A construction foreman exerts 1300 Newtons of force trying to move a 1200-kg block of concrete. How many Joules of work does he perform?

Answers

He does no work until the block starts to move ... an unlikely event.

If it ever does move, then the work he does is

(1300) x (the distance the block moves, in meters) .

The unit is 'joules.

The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.A. 1 cm/s2
B. 2 cm/s2
C. 5 cm/s2
D. 6 cm/s2

Answers

The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.

What is the acceleration due to gravity?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².

The formula for the acceleration due to gravity is g=GM/r².

According to the question, the given values are :

Mass, M = 4 × 1016 kg or

M = 4 × 10¹⁶.

Radius, r = 16 km or,

r = 16000 meter.

G = 6.67 × 10⁻¹¹ Nm²/kg²

g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²

g = 0.0104 m/s² or,

g = 1.04 cm/s².

Hence, the acceleration due to gravity will be 1.04 m/s²

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Answer:

1 cm/s²

Explanation:

I just took the quiz

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$

As the spheres are released v₀ = 0

$V_f^2=2gH$

$V_f=√(2gH)$

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

$V_(10)=\sqrt2gH$

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and $V_(20)=-\sqrt2gH=-V_h$

Large sphere m₁ and $V_(10)=√(2gH)=V_h$

Before crash

$P_o=m_1V_(10)+m_2V_(20)$

After the crash

$P_f=m_1V_(1f)+m_2V_(2f)$

$P_o=P_f$

$m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)$

The conservation of kinetic energy

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$

$K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

$K_o=K_f$

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$ $=$ $K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

Let's write the values

$-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)$

$m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2$

The solution to this system of equations is

$m_t=m_1+m_2$

$V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)$

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)$

The large sphere is labeled 1, we are asked for the mass so that $V_(1f)$ = 0, let's clear the equation

$V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))$

$0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))$

$(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)$

$(m_1-m_2)=2m_2$

$m_1=3m_2$

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))$

$V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$ mt = 3m2 + m2

$m_t=4m_2$

$V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)$

$V_(2f)=(3)/(2) V_h+(1)/(2) V_h$

$V_(2f)=2V_h$

$V_(2f)=2\sqrt{2gh$

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$ = 0

$V^2=V_(2f)^2-2gy$

$0=(2√(2gh))^2-2gy$

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

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Answer:

a) the large sphere has 3 times the mass of the small sphere

b) y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

vf² = vo² - 2g y

As the spheres are released v₀ = 0

vf² = - 2g H

vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

pf = m₁ v₁f + m₂ v₂f

po = pf

m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

Ko = KF

½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

mt = m₁ + m₂

v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

(m₁-m₂) / mt vh = 2 m₂ / mt vh

(m₁-m₂) = 2m₂

m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

mt = 3m2 + m2

mt= 4m2

v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

v₂f = 3/2 vh +1/2 vh

v₂f = 2 vh

v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

V² = v₂f² - 2 g Y

0 = (2√2gh)² - 2gy

2gy = 4 (2gH)

y = 4H

When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.

Answers

Answer:

P = 1.99 10⁸ Pa

Explanation:

The definition of the bulk module is

B = - P / (ΔV / V)

The negative sign is included for which balk module is positive, P is the pressure and V that volume

They tell us that the variation in volume is 9.05%, that is

ΔV / V = 9.0Δ5 / 100 = 0.0905

P = - B DV / V

P = 2.2 10⁹ (0.0905)

P = 1.99 10⁸ Pa

What is an atomic nucleus

Answers

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

Explanation:

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