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4 years ago

Why do we accelerate forward when we run?

1 answer:

8 0

It depends on what direction you are running in, if it's forward, you'll have to keep to accelerating forward to keep running.

If you stop accelerating forward, you wouldn't be able to move forward.

You might be interested in

To convert minutes per second into kilometre per hour we multiply the speed with

ankoles [38]

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

please mark as brainliest

3 0

3 years ago

A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23 m/s. If the ball was caught 42 m from the

ExtremeBDS [4]

Answer:

a) $v_{0y} = 9.72 m/s$

b) $v_{0x} = 20.85 m/s$

c) $t = 2.01 s$

d) $h_{max} = 4.82 m$

Explanation:

a) The initial vertical velocity is given by:

$v_{0y} = v*sin(\theta)$

Where:

θ: 25°

v: is the magnitude of the speed = 23 m/s

$v_{0y} = 23 m/s*sin(25) = 9.72 m/s$

b) The initial horizontal velocity can be calculated as follows:

$v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s$

c) The flight time can be calculated using the following equation:

$v_{0x} = \frac{x}{t}$

Where:

x: is the total distance = 42 m

$t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s$

d) The maximum height is given by:

$v_{fy}^{2} = v_{0y}^{2} - 2gh_{max}$

Where:

$v_{fy}$ : is the final vertical velocity =0 (at the maximum heigth)

g: is the gravity = 9.81 m/s²

$h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m$

I hope it helps you!

7 0

3 years ago

An unmanned vehicle to explore the solar system

grin007 [14]

Space ships.

Hope this helps and have a nice day!!!

4 0

3 years ago

A 1,000 kg ball traveling at 5 m/s would have

jonny [76]

Answer:

15 because 5×5×5 is the same thing as 5×3 which equals to 15

6 0

3 years ago

Read 2 more answers

In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a proton, where the radius of the orbit is

Alchen [17]

Answer:

(a) Electric force will be $77.285\times 10^{-9}N$

(B) Velocity $v=2.1522\times 10^6m/sec$

Explanation:

We have given that radius of the circular orbit $r=0.546\times 10^{-10}m$

Charge on electron $e=1.6\times 10^{-19}C$

(A) According to coulomb's law electric force between two charges is given by

$F=\frac{kq}{r^2}$

So electric force $F=\frac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.546\times 10^{-10})^2}=77.285\times 10^{-9}N$

We know that centripetal force is given by $F=\frac{mv^2}{r}$

So $\frac{9.11\times 10^{-31}\times v^2}{0.546\times 10^{-10}}=77.285\times 10^{-9}$

$v=2.1522\times 10^6m/sec$

8 0

3 years ago