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2 years ago

What is Marie's instantaneous speed at 20 minutes in miles/min?

Physics

1 answer:

AVprozaik [17]2 years ago

8 0

Answer:

0.25miles/min

Explanation:

Instantaneous speed of a person or an object is its speed at a particular moment usually at a period of time.

The speedometer of a car reports the instantaneous speed.

It can be mathematically expressed as;

Instantaneous speed = $\frac{distance}{time}$

At 20min the distance covered is 5miles;

Instantaneous speed = $\frac{5 miles }{20mins}$ = 0.25miles/min

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please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ba

viva [34]

Answer:

the acceleration during the collision is: - 5 $\frac{m}{s^2}$

Explanation:

Using the formula:

$a=\frac{\Delta\,v}{\Delta\,t}$

we get:

$a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}$

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3 years ago

1. What affects a material's resistance?

Ghella [55]

Answer:

A. Thickness and temperature

Explanation:

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3 years ago

A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is

Anika [276]

Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

Explanation:

For 2 quantities A and B represented as

$A\pm \Delta A$ and $B\pm \Delta B$

The sum is represented as

$Sum=(A+B)\pm (\Delta A+\Delta B)$

For the the values given to us the sum is calculated as

$Sum=(2.9+3.9)\pm (0.1+0.2)$

$Sum=6.8\pm 0.3$

Now the since the uncertainity inthe sum is $\pm 0.3$

The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

Thus closest distance equals $6.8-0.3=6.5$ meters

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3 years ago

A student throws a ball upward with a velocity of 35 m/s. What is the acceleration of the ball as it rises to the top of its arc

lora16 [44]

Answer:

The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

Explanation:

Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.

Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.

Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

7 0

2 years ago

A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co

Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W$

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = $-F_Dd$

Where:

d = Distance traveled = 450 m

Therefore,

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd$ and

$v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh -F_Dd}{ \frac{1}{2}m} = v_1^2 + 2gh -\frac{ 2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}$

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

6 0

2 years ago