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babunello [35]
3 years ago
12

At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the oppo

site direction. What was the average acceleration of the particle during this 6.0 s interval?
Physics
1 answer:
Viefleur [7K]3 years ago
8 0

Answer:

The average acceleration during the 6.0 s interval was -27 m/s².

Explanation:

Hi there!

The average acceleration is defined as the change in velocity over time:

a = Δv/t

Where:

a = acceleration.

Δv = change in velocity = final velocity - initial velocity

t = elapsed time

The change in velocity will be:

Δv = final velocity - initial velocity

Δv = -74 m/s - 87 m/s = -161 m/s

(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)

Then the average acceleration will be:

a = Δv/t

a = -161 m/s / 6.0 s

a = -27 m/s²

The average acceleration during the 6.0 s interval was -27 m/s².

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An electric current of 0.25 A passes through a circuit that has a resistance of
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Answer:

The voltage is V = 37.5 [V]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.

V = I*R

where:

V = voltage [Volt]

I = current = 0.25[amp]

R = resistance = 150 [ohm]

V = 0.25*150 = 37.5 [V]

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13. Which compound is most reactive?<br> ethene<br> ethyne<br> ethane
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Ethyne consists of a triple bond that is most easier to be broken as compared to that of ethyne with a double bond and benzene single bonds. Therefore ethyne is most reactive among the given options.

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2 years ago
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A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
galina1969 [7]

Answer:

\mu_s = 0.62

\mu_k = 0.415

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Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

F = ma\\F = 1.7m

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

5 0
3 years ago
How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?
Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations.  We are going to use vf = vi + at.  Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.  

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(0.5)×(0squared)×(3)=(1.5j)
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