A boat can travel 2.30 m/s in still water. If the boat points its prow directly across a stream whose current is 1.80 m/s , what is the magnitude of the velocity of the boat relative to the shore?

Answer:

Definition: Pressure:- Force per unit area surface is called pressure. P = F/ A Pressure = Force /Area Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. Gauge pressure is the pressure relative to the ambient pressure. 

Answer:

A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force.

Pressure is defined to be the amount of force exerted per area.

Formula of pressure = P = F÷A

Where, P = Pressure,

            F = Force applied and,

            A = Unit area.

Examples of Pressure is Air pressure.

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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

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In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

_________________________________

If gravity acceleration is 10 :

svf = 10 × 10 = 100 N

So ;

frictional force = 100 × 0.20

frictional force = 20 N

##############################

If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

So ;

frictional force = 98 × 0.20

frictional force = 19.6 N

_________________________________

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Answer:

A.) acceleration= 55.6m/s^2

B.) acceleration of table= 5.0m/s^2

C.) More acceleration

Explanation:

A.) 100N/1.8kg= -55.6

B.) 100N÷20kg= 5

C.) Because since the table would have less mass, it would have had to accelerate more

The acceleration of Claudia's foot is 55.56 m/s². The acceleration of the table is 5 m/s².  If the table had less mass then its acceleration will be less and Claudia's toe hurt less.

Acceleration can be described as the rate of change of velocity with respect to time. Acceleration is a vector quantity with both magnitude and direction. Acceleration is also the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.

According to the second law of Newton, the force is equal to the product of mass and acceleration.

F = ma

or, a = F/m

Given, the force with which Claudia stubs her toe on the coffee table,

F = 100N

The mass of Claudia's foot = 1.80 Kg

The acceleration of the Claudia's foot = 100/1.50 = 55.56m/s²

The mass of the table = 20 Kg

The acceleration of the table = 100/20 = 5 m/s²

The acceleration of any object is inversely proportional to its mass. If the table had less mass then its acceleration will be less and it would less hurt Claudia's toe.

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Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

where,

F = Force between Romeo and Juliet = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Romeo = 68 kg

m₂ = mass of space station = 4.58 x 10⁵ kg

r = distance = 25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

F =  3.32 x 10⁻⁶ N

Answer: This is Yazan From 10-C

Explanation

Answer:

speed = 72.75km/hr

Explanation:

Let us first determine the Kinetic energy (KE) of the 18000kg truck

[tex]KE = \frac{1}{2} mv^2\\where:\\m = mass = 18000\\v = velocity = 21km/hr\\\therefore KE = \frac{1}{2} \times 18000 \times (21)^2\\= 9000 \times 441\\KE = 3,969,000\ Joules\\= 3,969\ KJ[/tex]

Next we will substitute this value of kinetic energy into the KE equation for the 1500kg compact car

[tex]KE = \frac{1}{2} mv^2\\3969000 = \frac{1}{2} \times 1500 \times v^2\\3969000 = 750 \times v^2\\v^2 = \frac{3969000}{750} = 5292\\v = \sqrt{5292} \\v = 72.75km/hr[/tex]

The speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h

The kinetic energy of an object is known to be the energy at work and in motion. It is usually expressed as;

[tex]\mathbf{K.E = \dfrac{1}{2}mv^2}[/tex]

To determine the speed at which the car will have the same kinetic energy as the truck, we can say that:

[tex]\mathbf{\implies \dfrac{1}{2}m_c v_c^2= \dfrac{1}{2}m_t v_t^2}[/tex]

∴

[tex]\mathbf{\implies \dfrac{1}{2}\times 1500\times v_c^2= \dfrac{1}{2}\times 18000 \times 21^2}[/tex]

[tex]\mathbf{v_c^2= \dfrac{3969000}{ 750}}[/tex]

[tex]\mathbf{v_c^2=5292}[/tex]

[tex]\mathbf{v_c=\sqrt{5292}}[/tex]

[tex]\mathbf{v_c\simeq 73 km/h}[/tex]

Therefore, we can conclude that the speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h

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We are given:

Homer the Astronaut:

Mass of Homer the astronaut(m1) = 200 kg

initial velocity of Homer the astronaut(u1) = 5 m/s

Larry the Pal:

Mass of Larry the Pal (m2) = 150 kg

initial velocity of Larry the Pal (u2) = 0 m/s

Since they will move together after the collision, they will have the same velocity:

v1 = v2 = V

Solving for the Final velocity:

from the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

since v1 = v2 = V:

m1u1 + m2u2 = V(m1 + m2)

replacing the variables with the given values

200 * 5 + 150 * 0 = V(200 + 150)

1000 = 350V

V  = 1000 / 350

V = 2.86 m/s

The force required to hold it completely submerged under water is 0.252 N

As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.

Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation

Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m

Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³

Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg

Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N

Volume of water displaced = 4/3 π r³ = 2.805 e-5

Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N

F = 0.275 - 0.023 = 0.252 N

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The force required to hold it completely submerged under water is 0.25 N

The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³

The density of water [tex]\rho_w[/tex] = 1000 kg/m³

Diameter = 3.77 cm = 0.0377 m

radius of ball = 0.0377/2 = 0.01885 m

The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]

Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:

The force is required to hold it completely submerged under water (F) is:

[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]

F = 0.25 N

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Answer:

(a) V₂ = 1.86 m/s

(b) Q = 5.1 x 10⁻⁴ m³/s

Explanation:

(a)

The formula derived for Venturi tube is as follows:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa

ρ = Density of Gasoline = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

Therefore,

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²   ------------------- equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

Therefore,

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

using this value in equation (1):

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

(b)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)

Q = 5.1 x 10⁻⁴ m³/s

The formula for finding variables in a  Venturi tube is shown below:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity

P₁ - P₂ = 1.2 KPa = 1200 Pa

ρ = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

We then substitute the variables into this equation.

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²  ------ equation (1)

The  continuity equation A₁V₁ = A₂V₂ is then used

where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

We then substitute the value into equation 1

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

This can be calculated using the formula

Flow Rate = Q = A₂V₂

                  = (2.746 x 10⁻⁴ m²)(1.86 m/s)

                 = 5.1 x 10⁻⁴ m³/s

Solution:

The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :

[tex]$E=\frac{V}{D}$[/tex]

Differentiating on both the sides with respect to time, we get

[tex]$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$[/tex]

Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :

[tex]$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$[/tex]

      [tex]$=\frac{1}{1.4 \times 10^{-3}} \times 110$[/tex]

      [tex]$=7.85 \times 10^4$[/tex]  V/m-s

Answer:

they are cooler than the rest if the sun

Answer:

D- Z, X, Y, W

Explanation:

Correct on edge

Answer:

What is the normal force on the object (draw a diagram if needed)

784 N

How much force is required to get the object to start to move from rest (max static friction)?

  470.4 N

When the object starts moving, what is the force of kinetic friction?

274.4 N

If the moving object has a tension force of 300N to the right pulling it, what is the net force on the object in the horizontal direction? 784 N

What is the acceleration (with direction) of the object based on your answer for part d?  Remember that a = net force/m.

Explanation:

We have that for the Question "" it can be said that  the normal force and How much force is required to get the object to start to move from rest is

From the question we are told

An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface

a)

Generally the equation for the Normal Force  is mathematically given as

[tex]F_N=mg\\\\F_N=80*9.8\\\\F_N=784N\\\\[/tex]

b)

[tex]F_{max}=0.60*800\\\\\F_{max}480N\\\\[/tex]

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Complete Question

An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface

Answer:

yeah

Explanation:

yeah yeah yeah yeah

Answer:

work

Explanation:

Answer:

A

Explanation:

Answer:

The total energy is 53 Joule

Explanation:

Mechanical Energy

The mechanical energy of an object of mass m, speed v, and at a height h is:

[tex]\displaystyle E = m.g.h+\frac{mv^2}{2}[/tex]

The ball has a mass of m=0.5 Kg, a speed v = 4 m/s, and at a height of h=10 m. Thus the total energy is:

[tex]\displaystyle E = 0.5\cdot 9.8\cdot 10+\frac{0.5\cdot 4^2}{2}[/tex]

E = 49 J + 4 J = 53 J

The total energy is 53 Joule

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Answer:

leon

Explanation:

leom ofjfjbfbfdnns

It is the proportion predicted to be present in the early universe.

The hydrogen and helium abundance helps us to model the expansion rate of the early universe.

In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.

In big Bang model, the universe is mostly light or protons.

The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.

Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all  it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.

Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.

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Answer:

Brittleness

Explanation:

Lustrous means shiny

sonorous means capable of producing a deep or ringing sound

and iron isn't brittle or weak

Answer:

[tex]\huge\boxed{\sf brittleness}[/tex]

Explanation:

Iron is a metal which have luster (shine) so it is lustrous (shiny).

Iron is sonorous. When it is hit with a hard object, it produces ringing sounds.

Iron is not brittle. It cannot be easily broken. So, Iron does not show brittleness (It cannot be easily broken)

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

Answer:

b

Explanation:

i took the test

Earth’s landmasses were first recognized as the continents we know today

135 million years ago.

Landmass is defines as a large area of land. It can also be referred to as the

continents we have today. There are seven types of earth landmasses and

they are

They were first discovered around 135 million years ago by the early

dwellers of the earth.

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Answer:

0.78m/s^2

Explanation:

using the third equation of motion, the answer is found

v^2=u^2 -2gs

where V= 0 at the maximum point

2gs=u^2

g=u^2

2s

g=2.5^2

2(4)

g=0.78125m/s^2

or g=0.78m/s^2

Answer: Work done is  - 501.56 J

Explanation:

Given that;

External pressure P = 15.0 atm

Volume V1 = 0.120 liters

Volume V2 = 0.450 liters

Work done = ?

we know that; Work = -Pdv

where P is pressure and dv is change in volume

so we substitute our values into the equation

Work = -15.0 × ( 0.450 - 0.120)

= -15 × 0.33

= - 4.95 atm/L

we know that;

1 atm.L = 101.325 J

so

- 4.95 atm/L = 101.325 J × -4.95 atm/L ÷

= - 501.56 J

Therefore Work done is  - 501.56 J

Answer:

[tex]\mid \vec C\mid=31.9[/tex]

Explanation:

Consider the vectors:

[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]

[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]

Calculate the magnitude of

[tex]\vec C=-2\vec B-\vec A[/tex]

Substitute the values of the vectors:

[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]

Operate and remove parentheses:

[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]

Operating both components separately:

[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]

Now find the magnitude of C:

[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]

[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]

[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]

The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration a of

a = (0 - 1.3 m/s) / (0.30 s)

a ≈ -4.33 m/s²

This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass m such that

-65 N = m (-4.33 m/s²)

m = 15 kg