Which of the following sketches represents a possible configuration for this problem?

As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward

juin [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v =0.333 \ m/s$ in positive x -direction

The speed is $s = 0.733 \ m/s$

Explanation:

From the question we are told that

The distance from the house to truck is D = 20 m

The distance traveled back to retrieve wind-blown hat is d = 15

The distance from the wind-blown hat position too the truck is k = 20 m

The total time taken is t = 75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

$L = 20 - 15 + 20$

=> $L = 25 \ m$

Generally the average velocity is mathematically represented as

$v = \frac{L}{t}$

=> $v = \frac{25}{75}$

=> $v =0.333 \ m/s$

Generally the distance covered by Justin is mathematically represented as

$R = D+ d + k$

=> $R = 20 + 15 +20$

=> $R = 55 \ m$

Generally Justin's average speed over a 75 s period is mathematically represented as

$s = \frac{R}{ t}$

=> $s = \frac{55}{ 75}$

=> $s = 0.733 \ m/s$

8 0

2 years ago

At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w

aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s
Momentum of car A + B = 15,000 x 2m/s

So combining the two, will give us the equation:
15,000/5,000 = 3
3 x 2 =6m/s

6 0

2 years ago

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm). at the top of the circle the spee

gayaneshka [121]

Answer:

Tension in the string at this position: 3.1 N.

Explanation:

Convert the radius of the circle to meters:

$r = 80\;\text{cm} = 0.80\;\text{m}$ .

What's the net force on the object?

The object is in a circular motion. As a result,

$\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r}$ ,

where

$\Sigma F$ is the net force on the object,
$m$ is the mass of the object,
$v$ is the velocity of the object, and
$r$ is the radius of the circular motion.

For this object,

$\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}$ .

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

Weight, $W$ , which points downwards. $W = m\cdot g = 0.20\times 9.81 = 1.962\;\text{N}$ .
Tension, $T$ , which also points downwards. The size of the tension force needs to be found.

What's the size of the tension force?

Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,

$\Sigma F = T + W$ .

$T = \Sigma F - W = 5.0625 - 1.962 = 3.1$ .

All three values in this question are given with two sig. fig. Round the value of $T$ to the same number of significant figures.

4 0

2 years ago

A boy is standing motionless on a skateboard. He throws a basketball forward. Describe his motion.

weqwewe [10]

Answer:

Random motion

Explanation:

If the boy throws the basketball forward while at a position on the skateboard, the motion of the ball will be a random motion since we are not told if the ball is moving on a straight line when thrown forward.

In this case, the boy will tend to move in the direction of the ball. Since the ball is moving in a random manner, the motion of the boy will also be a random motion.

A random motion is a motion of a body in a zig zag manner. It is also known as Brownian motion e.g motion of a buzzing mosquito, motion of a smoke coming out of a chimney etc.

8 0

2 years ago

The brake in most cars makes use of a hydraulic system. This system consists of a fluid filled tube connected at each end to a p

Maksim231197 [3]

Answer:

The force at the brake pad = 250 N

Explanation:

The hydraulic brake system works on the Pascal's Principle for pressure transmission in fluids; the pressure applied to a fluid is transmitted undiminished in all directions.

For hydraulic systems, the pressure applied to the brake pedal is transmitted undiminished through the fluid filled tube, connected at each end to a piston, to the brake pad.

Hence, mathematically,

P(brake pedal) = P(break pad)

Pressure is given as the force applied divided by the cross sectional Area perpendicular to the direction of applied force.

P(brake pedal) = (Force applied on the brake pedal) ÷ (Cross Sectional Area of the brake pedal)

Force applied on the brake pedal = 50 N

Cross Sectional Area of the brake pedal = 3 cm²

P(brake pedal) = (50/3) = 16.67 N/cm²

P(brake pad) = P(brake pedal) = 16.67 N/cm²

P(brake pad) = (Force applied on the brake pad) ÷ (Cross Sectional Area of the brake pad)

Force applied on the brake pad = F = ?

Cross Sectional Area of the brake pad = 15 cm²

16.67 = (F/15)

F = 16.67 × 15 = 250 N

Hence, the force at the brake pad = 250 N

Hope this Helps!!!

7 0

2 years ago