All categories

3 years ago

4. If you travel the first half of a trip at 20 mi/h, how fast must you travel the second half of the trip so that your

Physics

1 answer:

iVinArrow [24]3 years ago

5 0

Answer:

60 mi/h

Explanation:

An average is:

$m = \frac{\text{sum of the terms}}{\text{number of terms}}$

And this is the same in physics.

So:

20+60 =80

80/2= 40

You might be interested in

Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they

maw [93]

This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:

Ff = UsN

where Us is the coefficient of static friction and N is the normal force.

In order to get the crate moving you must first apply enough force to overcome the static friction:

Fapplied = Ff

Since Fapplied = 43 Newtons:

Fapplied = Ff = 43 = UsN

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2

7 0

3 years ago

Read 2 more answers

at a drag race, the light turns green and 0.00125 hours later, a dragster is travelling 300 miles per hour. Calculate the accele

gtnhenbr [62]

240,000 miles / hourÂ˛ Average acceleration can be calculated by dividing the change in speed by the elapsed time. Since the dragster's speed was 0 when the light turned green, the change in speed is simply 300 mph. Now, divide that by the time: 300 mph / 0.00125 hours = 240,000 miles / hourÂ˛ By the way, 0.00125 hours is just 4.5 seconds!

5 0

4 years ago

Read 2 more answers

A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b

Artist 52 [7]

Answer:

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

Distance traveled during time t = 4 cm

Initial velocity = u m/s

Acceleration = 0 $m/s^2$

So $4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}$

Now at time 2t,

$S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm$

So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

Distance traveled during time t = 4 cm

Initial velocity = 0 m/s

Acceleration = -g $m/s^2$

$4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}$

At time 2t,

$S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm$

So vertical distance traveled in time 2t = 16 cm to the bottom

6 0

3 years ago

A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du

Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}$ .

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is $700\,{\text{kg}}$ .

The velocity of the car before collision is $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The velocity of the car after collision is $4.5\text{ m}/\text{s}$ .

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

$\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}$ …… (I)

Here, ${p_i}$ is the initial momentum of the car, $m$ is the mass of the car and ${v_i}$ is the initial velocity of the car.

Substitute $700\,{\text{kg}}$ for $m$ and $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (I).

$\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}$

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

$\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}$ …… (II)

Here, ${p_f}$ is the final momentum and ${v_f}$ is the final velocity.

Substitute $700\,{\text{kg}}$ for $m$ and - $4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_f}$ in equation (II).

$\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

$\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}$ …… (III)

Here, $\Delta p$ is the change in the momentum of the car.

Substitute $- 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${p_f}$ and $20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (III).

$\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}$

Thus, the change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}$ .

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.