All categories

3 years ago

A person walks briskly at 1.90 m/s .how much time will it take them to walk one mile?

Physics

1 answer:

Paladinen [302]3 years ago

5 0

As in question, the speed of person is given m/s whereas the distance he needs to cover is given in miles.

Formula to calculate the time,

$Distance = velocity \times time$

To get the correct answer, convert the distance in to miles.

$1 mile = (1\times1.6093)\times1,000=1,609.3$

Therefore,

$1,609.3 meters = 1.90 m/s \times time$

$time = \frac{1,609.3 m}{1.90 m/s}$

$time = 847 s$

You might be interested in

A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc

vredina [299]

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 6,000 kg$ is the mass of the first train

$u_1 = 10 m/s$ is the initial velocity of the first train

$m_2 = 4,000 kg$ is the mass of the second train

$u_2 = 0$ is the initial velocity of the second train (initially at rest)

$v$ is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s$

There are two types of collisions:

Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

$K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J$

After:

$K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J$

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

3 0

3 years ago

A ? watcher is a citizen who is paid by the parties to keep a watchful eye on the voters and the officials.

VARVARA [1.3K]

A poll watcher
<span>a person who is paid by the parties to watch the voters and officials to make sure everything is fair</span>

6 0

3 years ago

Read 2 more answers

The formula to calculate velocity is

SashulF [63]

speed = distance/time

7 0

3 years ago

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str

Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, $v_y$ , of the package just before landing is given by the relation;

$v_y$ ² = u² + 2·g·h

u = 0 m/s

∴ $v_y$ ² = 0 + 2×9.81×1500 = 29430 m²/s²

$v_y$ = √29430 = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

$tan \theta = \dfrac{v_y}{v_x} = \dfrac{171.55}{39.0 } = 4.4$

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0

3 years ago