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mihalych1998 [28]
3 years ago
13

Man is walking due east at the rate of of 4kmph and the rain is falling 30° east of vertical with a velocity of 6kmph the veloci

ty of rain relative to the man will be?
​
Physics
1 answer:
Tresset [83]3 years ago
7 0

Answer:

No answer

Explanation:

no explanation

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A 70 kg bicyclist rides his 9.8 kg bicycle with a speed
Sliva [168]
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

<em>✔ We have Ek = 1/2m x v² </em>

  • Ek = 1/2 x 79.8 x 16²
  • Ek = 10 214.4 J

In order to come to a complete stop, the cyclist must convert all his kinetic energy into thermal energy. Given that the braking force opposes movement, the work is therefore resistant, i.e.  W = -10 214.4 J.

8 0
3 years ago
The velocity of a projectile at launch has a horizontal component vh and a vertical component vv. When the projectile is at the
Usimov [2.4K]

Answer:

- horizontal component of the velocity: v_h (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: -g=-9.8 m/s^2 (downward)

Explanation:

The motion of a projectile consists of two independent motions:

- Along the horizontal direction, there are no forces acting on the projectile (if we neglect air resistance), therefore the horizontal acceleration is zero and the horizontal component of the velocity, vh, is constant

- Along the vertical direction, there is only one force acting on the projectile: the force of gravity, downward, which produces a constant downward acceleration of g=9.8 m/s^2. As a consequence, the vertical component of the velocity changes according to

v_v(t) = v_v-gt

where vv is the initial vertical velocity and t the time. According to this equation, the vertical component of the velocity decreases first, then becomes zero at the point of maximum height, then becomes negative (= changes direction and points downward)

So, in summary, at the highest point of the trajectory we have:

- horizontal component of the velocity: v_h (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: -g=-9.8 m/s^2 (downward)

3 0
3 years ago
Are the correct? PLEASE HELP ASAP
Trava [24]

Energy is the ablity to dow work.

Power is the time rate at which work is done.

independent variable often along x axis, dependent along y axis.

You can set the independent value, from which the dependent follows.


7 0
3 years ago
Read 2 more answers
Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.
klasskru [66]

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

a = -36.81 m/s²

The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

4 0
3 years ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
3 years ago
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