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Nutka1998 [239]

2 years ago

12

A train is moving with a velocity of 20 m/s. when the brakes are applied the acceleration is reduced to -0.6 m/s².calculate the

distance covered by the train before coming to rest

1 answer:

Serhud [2]2 years ago

3 0

Answer:

x = 333.33 [m]

Explanation:

To solve this problem we must use the following kinematics equation.

$v_{f}^{2} = v_{i}^{2}-(2*a*x)$

where:

Vf = final velocity = 0

Vi = initial velocity = 20 [m/s]

a = desacceleration = 0.6 [m/s^2]

x = distance [m]

Note: the final speed is zero as the body finishes its movement.

Now replacing:

0 = (20)^2 - (2*0.6*x)

1.2*x = 400

x = 333.33 [m]

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With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

#A

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

#B

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

#C

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

4 0

1 year ago

Simple but not plagiarized answer for "what is forces and motion?"

egoroff_w [7]

FORCE-physical power or strength possessed by a living being:
He used all his force in opening the window.

MOTION-the action or process of moving or of changing place or position; movement.

7 0

2 years ago

The speed of sound in air is approximately 340 m/s. The speed of light in air is approximately 3 x 108 m/s. If 10 seconds elapse

olya-2409 [2.1K]

Answer:

3400 m

Explanation:

Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;

speed = distance/time, hence, distance = speed x time.

<em>For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as</em>;

distance = 340 m/s x 10 s = 3400 m

3 0

2 years ago

Which of the following are true statements about the electric field? a. A moving charge produces a magnet anomaly. b. The electr

mars1129 [50]

Answer:

a,b,c,d,,f, g, j

Explanation:

e) equipotential lines are lines connecting points of equal potential

h) electric field inside the conductor is non-zero even when there is net movement of charge or non-zero current.

i) capacitors' plates are charged and an electric field exists between the plates.

4 0

2 years ago

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

3 0

2 years ago