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2 years ago

A train moves from rest to a speed of 25 m/s in 50.0 seconds. What is its acceleration?

Physics

1 answer:

ELEN [110]2 years ago

5 0

Answer:

Wellll. I am assuming the direction of speed is in the same direction as the direction of displacement of the train. (i.e. Velocity is positive)

Acceleration is defined as the rate of change of velocity with respect to time (m^s-2)

Explanation:

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Which equation describes the line containing the points (-2, 3) and (1, 2)

snow_lady [41]

Answer:

$y = \frac{ - 1}{3}x + \frac{7}{3}$

Explanation:

$\frac{y - 3}{2 - 3} = \frac{x + 2}{1 + 2} \\ \ - y + 3 = \frac{x + 2}{3} \\ y = \frac{ - 1}{3}x + \frac{7}{3}$

8 0

2 years ago

While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it

diamong [38]

Answer: 6s

Explanation:

Vs=32m/s speed at beginning of slowing down

Vf=0m/s stop speed

a= -6 m/s² acceleration

----------------

Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

t=6 s

7 0

2 years ago

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at

VARVARA [1.3K]

Answer:

μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

Before starting to move the box

Y axis

N-W = 0

N = W = mg

X axis

F -fr = 0

F = fr

The friction force has the formula

fr = μ N

fr = μ m g

At the limit point just before starting the movement

F = μ m g

μ = F / m g

calculate

μ = 34.8 / (19.8 9.8)

μ = 0.18

7 0

2 years ago

Read 2 more answers

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

Pls help i will be so happy thank you

Anvisha [2.4K]

Answer:

below given.

Explanation:

1) linear/direct

2) Indirect

3) linear/direct

4) quadratic

5) Inverse

6) linear/direct

7) Inverse / Indirect

8) Inverse / Indirect

4 0

2 years ago