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Nutka1998 [239]

3 years ago

10

(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the

Earth, assuming its speed relative to the water remains 7.00 m/s ? (b) What would its speed be relative to the Earth?

1 answer:

shtirl [24]3 years ago

4 0

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

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A bag weighing 20 N is pushed horizontally a distance of 35 m across a

topjm [15]

Answer:

350J

Explanation:

Given parameters:

Weight of bag = 20N

Distance moved horizontally = 35m

Force applied = 10N

Unknown:

Work done on the bag = ?

Solution:

Work done is the force applied to move a body through given distance.

Work done = Force applied x distance

So;

Work done = 10 x 35 = 350J

6 0

3 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

$v_{min} \approx 17.153\,\frac{m}{s}$

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

The minimum velocity is:

$v_{min} = \sqrt{2\cdot g \cdot h}$

Let assume that radio of curvature is measured in meters. Hence:

$v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}$

$v_{min} \approx 17.153\,\frac{m}{s}$

8 0

3 years ago

A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?

nydimaria [60]

Answer:

2.4 m/s". 1

Explanation:

A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1

8 0

3 years ago

A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien

Neporo4naja [7]

Answer:

The magnetic field is $B = 8.20 *10^{-3} \ T$

Explanation:

From the question we are told that

The mass of the metal rod is $m = 0.12 \ kg$

The current on the rod is $I = 4.1 \ A$

The distance of separation(equivalent to length of the rod ) is $L = 6.3 \ m$

The coefficient of kinetic friction is $\mu_k = 0.18$

The kinetic frictional force is $F_k = 0.212 \ N$

The constant speed is $v = 5.1 \ m/s$

Generally the magnetic force on the rod is mathematically represented as

$F = B * I * L$

For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

$F_ k = B* I * L$

=> $B = \frac{F_k}{L * I }$

=> $B = \frac{0.212}{ 6.3 * 4.1 }$

=> $B = 8.20 *10^{-3} \ T$

7 0

3 years ago

The distribution circuit of a residential power line is operated at 2000 V rms. This voltage must be reduced to 240 V rms for us

forsale [732]

Answer:

$N_p\approx3958$

Explanation:

In an ideal transformer the relationship between the voltages is proportional to the ratio between the number of turns of the windings. Thus:

$\frac{V_p}{V_s} =\frac{N_p}{N_s}$

Where:

$V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil$

$V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil$

$N_p=Turns\hspace{3}on\hspace{3}primary\hspace{3}coil$

$N_s=Turns\hspace{3}on\hspace{3}secondary\hspace{3}coil$

So, solving for $N_p$

$N_p=N_s*\frac{V_p}{V_s} =475*\frac{2000}{240} =3958.333333\approx3958$

7 0

3 years ago