With speeds up to 90 miles per hour, what is the fastest sport on ice?.

A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie

cluponka [151]

Answer:

a) v₃ = 19.54 km, b) 70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

cos 45 = x₁ / V₁

sin 45 = y₁ / V₁

x₁ = v₁ cos 45

y₁ = v₁ sin 45

x₁ = 26 cos 45

y₁ = 26 sin 45

x₁ = 18.38 km

y₁ = 18.38 km

Vector 2 moves 45 km north

y₂ = 45 km

Unknown 3 vector

x3 =?

y3 =?

Vector Resulting 70 km north of the starting point

R_y = 70 km

we make the sum on each axis

X axis

Rₓ = x₁ + x₃

x₃ = Rₓ -x₁

x₃ = 0 - 18.38

x₃ = -18.38 km

Y Axis

R_y = y₁ + y₂ + y₃

y₃ = R_y - y₁ -y₂

y₃ = 70 -18.38 - 45

y₃ = 6.62 km

the vector of the third leg of the journey is

v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

v₃ = √ (18.38² + 6.62²)

v₃ = 19.54 km

to find the angle let's use trigonometry

tan θ = y₃ / x₃

θ = tan⁻¹ (y₃ / x₃)

θ = tan⁻¹ (6.62 / (- 18.38))

θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

θ’= 180 -19.8

θ’= 160.19º

I mean the address is

θ’’ = 90-19.8

θ = 70.2º

70.2º north-west

3 0

4 years ago

An electric immersion heater is put at the bottom of a large tank of water. The water next to the heater becomes warm

djyliett [7]

Answer:

when the water is heated with immersion heater, the water becomes less dense due to which the warm water rises up and the cooler water fills it's space.

4 0

3 years ago

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T

Maksim231197 [3]

Answer:

35

Explanation:

We are given that

Initial voltage, $V_1=120 V$

Final voltage, $V_2=5 V$

Number of tuns in primary coil of the transformer, $N_p=840$

Rms current, $I_{rms}=580mA=580\times 10^{-3} A$

$1 mA=10^{-3} A$

We have to find the number of turns are there on the secondary coil.

We know that

$\frac{N_s}{N_p}=\frac{V_2}{V_1}$

Using the formula

$\frac{N_s}{840}=\frac{5}{120}$

$N_s=\frac{5}{120}\times 840=35$

Hence, there are number of turns on the secondary coil=35

8 0

3 years ago

In a vacuum, which of the following statements about terminal velocity must be true?

maw [93]

Objects have the same terminal velocity as they do when they is air so option A is correct

8 0

3 years ago

A lumberjack (mass = 110 kg) is standing at rest on one end of a floating log (mass = 206 kg) that is also at rest. The lumberja

Lemur [1.5K]

Answer:

a). The velocity of the first log is -1.65 m/s.

(b). The velocity of the second log is 1.07 m/s.

Explanation:

Given that,

Mass of lumberjack M= 110 kg

Mass of log m= 206 kg

Final velocity = 3.09 m/s

(a). We need to calculate the velocity of the first log just before the lumberjack jumps off

Using conservation of momentum

$Mu_{1}+mu_{2}=Mv_{1}+mv$

Put the value into the formula

$0=110\times3.09+206v$

$v=-\dfrac{110\times3.09}{206}$

$v=-1.65\ m/s$

The velocity of the first log is -1.65 m/s.

(b). If the lumberjack comes to rest relative to the second log

We need to calculate the velocity of the second log

$(M+m)v=Mv_{1}$

$v=\dfrac{Mv_{1}}{M+m}$

Put the value into the formula

$v=\dfrac{110\times3.09}{110+206}$

$v=1.07\ m/s$

The velocity of the second log is 1.07 m/s.

Hence, (a). The velocity of the first log is -1.65 m/s.

(b). The velocity of the second log is 1.07 m/s.

4 0

3 years ago