How far does a car travel if is going 4.0 m/s and accelerates to 3.5m/s2 for 5s ?

a car dealer sells 4 makes of cars in 5 colors with either standard or automatic transmissions. how many variations does the dea

Anuta_ua [19.1K]

<u>The number of variations the dealership have is 40 cars</u>.

Data given;

The number of different product = 4
The number of transmission = 2 (standard or automatic)
The numbers of colors available = 5

<h3>Dealership variation </h3>

This is the total numbers of cars available in the dealer shop at the moment.

To get that, we simply multiply the numbers of colors available, the number of transmission and the makes of car available.

This is equal to

$4*2*5=40$

From the above calculation, we can say that <u>the variations the dealership offers is 40</u>

<u />

Learn more on combination here

brainly.com/question/10339517

7 0

3 years ago

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The

Rufina [12.5K]

Answer:

<h2>4.25m/s</h2><h2>E. None of the option is correct</h2>

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

4 0

4 years ago

An atomic nucleus is composed of A) protons. B) protons and neutrons. C) protons and electrons. D) protons, neutrons, and electr

In-s [12.5K]

Answer:

B) protons and neutrons.

Explanation:

The protons and neutrons are located in the nucleus of the atom and represent most of the 'mass' of the atom, that's their count that determine the 'mass' of an atom (like 12 for Carbon).

The electrons rotate around the nucleus and have a negligible mass.

7 0

3 years ago

A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity

zvonat [6]

Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is <u>3.92 m/s</u>.

Explanation:

Let's use <u>Newton's 2nd Law</u> to help solve this problem.

F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: $F_c=115 \ \text{N}$ .

The mass "m" is the <u>sum</u> of the man and woman's masses: $85+65= 150 \ \text{kg}$ .

The acceleration is the centripetal acceleration of the Ferris wheel: $a_c=\displaystyle \frac{v^2}{r}$ .

Let's write an equation and solve for "v", the tangential (linear) acceleration.

$\displaystyle 115=m(\frac{v^2}{r} )$
$\displaystyle 115 = (85+65)(\frac{v^2}{20})$
$\displaystyle 115=150(\frac{v^2}{20} )$
$.766667=\displaystyle(\frac{v^2}{20} )$
$15.\overline{3}=v^2$
$v=3.9158$

The Ferris wheel's tangential velocity is 3.92 m/s.

8 0

2 years ago

Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr

Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

$W = dP + dK\\$

Where

$dP$ = Change in Potential Energy

$dK$ = Change in Kinetic Energy

Change in Potential Energy

$P_{i} = mgh_{i}\\ P_{f} = mgh_{f}$

Where

$P_{i}$ = Initial Potential Energy

$P_{f}$ = Final Potential Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$h_{i}$ = Initial Height = 0

$h_{f}$ = Final Height = 30 m

Inputting the values we get the answer for $dP$

$dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000$

Change in Kinetic Energy

$K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2$

Where

$K_{i}$ = Initial Kinetic Energy

$K_{f}$ = Final Kinetic Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$v_{i}$ = Initial Velocity = 0 m/2

$v_{f}$ = Final Velocity = 4 m/s

Inputting the values we get the answer for $dK$

$dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000$

Total Work

$W = dP + dK\\$

Inputting the values

$W = 2943000 + 80000$

$W = 3,023,000$

a) Finding the useful Power Output

$P = \frac{W}{t}$

Where

$P$ = Power Output

$W$ = Work Done = 3,023,000J

$t$ = Time = 12s

Inputting the values

$P = \frac{3,023,000}{12}\\ P = 251,916.667$

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0

3 years ago