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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?
Physics
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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Strike441 [17]
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= 20m/s
Velocity= 20m/s
In 4seconds,distance covered=20x4= 80m
Therefore,an accident occurs because in 4seconds, the car would have moved past 20m
6 0
2 years ago
If a 100 N net force acts on a 50 kg car, what will the acceleration of the car will be?
Allushta [10]
According to newton's law Force = mass * acceleration
so , 100 = 50 * a
so , a= 2 m/s^2
6 0
3 years ago
Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415
motikmotik

Answer:

35 N to the right.

Explanation:

450 is going to the right so you subtract what is going against it. Which gives you 35. And because 450 is bigger than 415, it'll be going to the right.

6 0
2 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it
makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

 Here, we have been given the height through which the ball drops;

 Height of drop = 14.5m - 1.9m  = 12.6m

The right motion equation is;

      V²  = U² + 2gh

V is the final velocity

U is the initial velocity  = 0

g is the acceleration due to gravity  = 9.8m/s²

h is the height

  Now insert the parameters and solve;

       V² = 0² + 2 x 9.8 x 12.6

      V²  = 246.96

       V = √246.96  = 15.7m/s

4 0
3 years ago
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