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Furkat [3]
2 years ago
9

Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

e riverbank sees the boats moving at 4.9 m/s and 7.0 m/s.(a) What is the speed of the boats relative to the river? (b) How fast is the river moving relative to the shore?
Physics
1 answer:
DENIUS [597]2 years ago
3 0

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

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Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
KIM [24]

Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

       t_1 =  250 \ s

The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

7 0
2 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to m
Umnica [9.8K]

Answer:

5

Explanation:

8 0
1 year ago
An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

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

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
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