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olchik [2.2K]
3 years ago
15

A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the

Physics
1 answer:
Elan Coil [88]3 years ago
7 0

Answer:

The horse is going at 12.72 m/s speed.

Explanation:

The initial speed of the horse (u) = 3 m/s

The acceleration of the horse (a)= 5 m/s^{2}

The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m

Applying the second equation of motion to find out the time,

s=ut+\frac{1}{2}at^{2}

15.3=3t+2.5t^{2}

2.5t^{2}+3t-15.3=0

Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.

Now applying first equation of motion, to find out the final velocity,

v=u+at

v=3+1.945*5

v=3+9.72

v=12.72 m/s

The horse travels at a speed of 12.72 m/s after covering the given distance.

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Suppose that a tiny space invader (m=0.47 kg) and the new space station orbit Earth at the same height from Earth’s surface. Whi
Ad libitum [116K]

The tiny space invader and the new space station will have equal orbital speed.

<h3>The orbital speed of satellite </h3>

The orbital speed of satellite on Earth surface is given as;

V = \sqrt{\frac{GM}{r} }

where;

  • V is the orbital speed
  • G is universal gravitation
  • M is mass of the Earth
  • r is the radius of the circular path

Since the two objects are at the same height from Earth's surface, the distance from the central point (r) will be the same. Thus, the two objects will have equal orbital speed.

Learn more about orbital speed here: brainly.com/question/22247460

6 0
2 years ago
K i Illeanalu plo,ppio<br>ng"i​
Ivenika [448]

whats that supposed to mean

6 0
2 years ago
Read 2 more answers
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
2 years ago
How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?
Anna [14]
Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.
7 0
2 years ago
Electromagnets can be used to lock doors. How could an engineer increase the strength of an electromagnetic lock? Check all that
ddd [48]

The answer should be A

3 0
3 years ago
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