The maximum amount of pulling force a truck can apply when driving on

In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second

lakkis [162]

If he keeps that pace he will be at the 34 yard line

7 0

2 years ago

Let’s tie this all together. It makes sense that, if the rope force remains greater than the gravitational force, the child keep

Pavlova-9 [17]

Answer:

If child weight is equal to rope force then child will move with uniform speed

or we can say that the child will remain at rest in his position

Explanation:

As we know that child is hanging by rope

so here there will be two forces on the child

1) Weight or gravitational force which act vertically downwards

2) Tension in the rope which act vertically upwards

Now if child will accelerate upwards then tension force must be more than the weight of the child

If tension force is less than the weight then child will decelerate and his speed will decrease

if tension force is equal to child weight then in that case the child will remain at rest or it will move with same speed

8 0

2 years ago

Which of the following can explains convection?

Vikentia [17]

Answer:

A) would be the best explanation because hot air is less dense and would rise to the top of a column of air

7 0

2 years ago

A motorcycle is moving at a speed of 21 m/s. The driver applies the brakes, giving hum an acceleration of -2.4 m/s^2. After 2 se

WARRIOR [948]

<h3><u>Answer;</u></h3>

B. 16.2 m/s

<h3><u>Explanation</u>;</h3>

Using the equation;

v = u + at; where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken;

u = 21 m/s, a= -2.4 m/s^2 and t = 2 seconds

Therefore;

v = 21 + ((-2.4) × 2)

= 21 - 4.8

<u> = 16.2 m/s</u>

6 0

2 years ago

In 0.601 s, a 13.1-kg block is pulled through a distance of 4.19 m on a frictionless horizontal surface, starting from rest. The

Naya [18.7K]

Answer:

0.615 m

Explanation:

We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.

The acceleration is determined using the equation of motion.

Given parameters:

Initial velocity, <em>u</em> = 0.00 m/s

Distance, <em>s</em> = 4.19 m

Time, <em>t</em> = 0.601 s

We use the equation

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

With <em>u</em> = 0.00 m/s,

$s = \frac{1}{2}at^2$

$a = \dfrac{2s}{t^2}$

$a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2$

The force is

$F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}$

From Hooke's law, the extension, <em>e</em>, of a string is given by

$e = \dfrac{F}{k}$

where <em>k</em> is the spring constant.

Hence,

$e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}$

4 0

3 years ago