A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats speed relative to the ground

Answers

Answer 1

Answer:

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

Related Questions

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?How far does the car move during the 2.52 s?How long does it take the car to come to a complete stop?
An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units
The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?
A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.
A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

Water flows at speed of 4.4 m/s through ahorizontal pipe of diameter 3.3 cm . The gauge
pressure P1 of the water in the pipe is 2 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.4 cm
(IMAGE)
What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 10^5 Pa . The density of water is 1000 kg/m^3
. The viscosity
of water is negligible.
Answer in units of atm

Answers

Answer:

1.75 atm

Explanation:

Mass is conserved, so the mass flow before the constriction equals the mass flow after the constriction.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Writing v₂ in terms of v₁:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ d₁²/d₂²)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (d₁/d₂)⁴

P₁ + ½ ρ v₁² (1 − (d₁/d₂)⁴) = P₂

Plugging in values:

P₂ = 2 atm + ½ (1000 kg/m³) (4.4 m/s)² (1 − (3.3 cm / 2.4 cm)⁴) (1 atm / 1.013×10⁵ Pa)

P₂ = 1.75 atm

The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. 1.11 × 104 Pa 1.09 × 105 Pa 1.13 × 107 Pa 1.11 × 108 Pa 2.18 × 105 Pa

Answers

Answer: 1.11 x 10⁸ Pa

Explanation:

At any deep, the absolute pressure is the same for all points located at the same level, and can be expressed as follows:

p = p₀ + δ. g . h, where p₀ = atmospheric pressure = 101, 325 Pa

Replacing by the values, we get:

p= 101,325 Pa + 1025 Kg/m³ . 9.8 m/s². 11,033 m = 1.11 x 10⁸ Pa.

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?

Answers

Answer:

The truck will reach there in 250 seconds.

Explanation:

The frequency due to doppler effect, when the observer is stationary and the source is moving towards it is

$f_(obv)$ = $(v)/(v-v_(s) ) f$

where v= velocity of sound in air

$v_(s)$ = velocity of source of sound

f= frequency of sound and

$f_(obv)$ = frequency oberved due to Doppler effect

$(v)/(v-v_(0) ) f$ = 460------------------------------------------( 1 )

The frequency due to doppler effect, when the observer is stationary and the source is moving away from it

$f_(obv)$ = $(v)/(v+v_(s) ) f$

where v= velocity of sound in air

$v_(s)$ = velocity of source of sound

f= frequency of sound and

$f_(obv)$ = frequency oberved due to Doppler effect

$(v)/(v+v_(0) ) f$ = 410-------------------------------------------( 2 )

Dividing ( 1 ) by ( 2 )

$(v+v_(s) )/(v-v_(s) ) =(460)/(410)$

$(v+v_(s) )/(v-v_(s) ) =(46)/(41)$

41v + 41 $v_(s)$ = 46v - 46 $v_(s)$

87 $v_(s)$ = 5v

$v_(s)$ = $(5)/(87)v$

Velocity of Sound (v)= 348 m/s

$v_(s)$ =20 m/s

Therefore, the truck is moving at 20 m/s.

$Time=(Distance)/(Time)$

Distance= 5000 m

Time= $(5000)/(20)$

Time= 250 s

Time = 4 min 10 sec

What is the force (in Newtons, 1 Newton = 1Kgm/s2) required to accelerate a 1500 Kg car to 3 m/s2?

Answers

Answer:

F=4500N

Explanation:

F=m×g

F=1500kg×3m/s²

F=4500N

Answer:

F=4500N

Explanation:

F=m×g

F=1500kg×3m/s²

F=4500N

The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.

Answers

Answer:

80 - 32t

Explanation:

The height, h, in terms of time, t, is given as:

h(t) = 650 + 80t − 16t²

Velocity is the derivative of distance with respect to time:

v(t) = dh(t)/dt = 80 - 32t

Final answer:

The velocity of the object as a function of time is given by the derivative of the height function, which is v(t) = 80 - 32t.

Explanation:

The height h(t) of an object is given by the equation h(t) = 650 + 80t − 16t2. To find the velocity v(t), we need to take the derivative of h(t) with respect to time t. Using the power rule, we get:

v(t) = dh/dt = 0 + 80 - 32t.

So, the velocity of the object as a function of time t is v(t) = 80 - 32t.

Learn more about Velocity Function here:

brainly.com/question/33157131

#SPJ11

Other Questions

"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 781.25 N when t = 1.27 s .What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?

URGENT!!!!!! Assume that a wire has 1.5 ohms of resistance. If the wire is connected to two batteries with a total voltage of 3.0 V, how much current will flow through the wire? 3.0 amps 2.3 amps 2.0 amps 1.0 amps

True Or False for each question

ou are writing a short adventure story for your English class. In your story, two submarines need to arrive at a place in the middle of the Atlantic Ocean at the same time. They start out at the same time from positions equally distant from the rendezvouspoint. They travel at different speeds, but both go in a straight line. The first submarine travels at an average speed of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. In the story’s plot, the second submarine is required to travel at a constant velocity, so the captain needs to determine the magnitude of that velocity. What is that velocity?