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2 years ago

13

A motor boat is traveling east with a velocity of 7.3 meter/seconds. Its experiences a current of 0.34 meters/seconds from the e

ast. Find the resultant velocity of the boat and its direction.

2 answers:

Delvig [45]2 years ago

6 0

Answer:

The resultant velocity of the boat and its direction is 6.96 m/s in east direction.

Explanation:

Given that,

Velocity of boat = 7.3 m/s in east direction

Velocity of current = 0.34 m/s in west from the east

The velocity of current is in opposite direction of the velocity of boat.

So, The resultant velocity of the boat will be

The resultant velocity = Velocity of boat - velocity of current

$R=7.3-0.34$

$R=6.96\ m/s$

The direction of the boat in east direction.

Hence, The resultant velocity of the boat and its direction is 6.96 m/s in east direction.

andreev551 [17]2 years ago

3 0

Since they're going in the same direction, just add the velocities together.

7.3 m/s + 0.34 m/s = 7.64m/s

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non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

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2 years ago

Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box, what is the magnit

vlabodo [156]

Newton's third law tells us that for every force there is an equal and opposite force. This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.

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2 years ago

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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

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2 years ago

An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a

wlad13 [49]

Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

$fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75$
$fcp = 66.666... = 66 \: newtons$

Answer: Last option, 66 N.

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2 years ago

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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

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= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

2 years ago