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balu736 [363]
3 years ago
12

A car traveling at 12 m/s slows down with an acceleration of -1.5 m/s2. What is the

Physics
1 answer:
Alex777 [14]3 years ago
7 0

Answer:

using s=ut +0.5at^2

u=12m/s

a=-1.5m/s2

t=6 s

s=54m

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A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?
SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0
3 years ago
The speed of propagation of electrical signals in a nerve cell depends on the diameter of the axon (nerve fiber). If the nerve c
gayaneshka [121]

Answer:

61 ms

Explanation:

Given that

Length of the nerve call connecting the spinal cord to your feet, s is 1.1 m

Speed of the impulse of the nerve, v = 18 m/s

Time taken by the nerve to travel the distance, t = ?

In solving this, we'd apply the very most basic formula in physics. Formula of speed, with respect to time and distance.

Speed, v = distance, s / time, t

time, t = distance, s / speed, v

time, t = 1.1 / 18

time, t = 0.061 s, converting to milliseconds, we have, time, t = 61 ms

6 0
3 years ago
BRAINLIESTTTT!!!!! THIS IS DUE IN 5 MINUTES!<br>​
Gnesinka [82]

Equation: -10 40/4^2

Explanation: I multiplied the -10, but idk if that' what you do cause I've never done this. So if you do do that it's -1000, but that is probably wrong. Sorry, I tried.

8 0
3 years ago
What is gravitational constant?<br>​
WITCHER [35]
The constant in Newton's law of gravitation relating gravity to the masses and separation of particles, equal to 6.67 × 10-11N m2 kg-2.
6 0
3 years ago
A small lead ball, attached to a 1.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
soldier1979 [14.2K]

Answer:

h = 57.6 m

Explanation:

First, we find the linear speed of the ball while in circular motion:

v = rω

where,

v = linear speed of ball = ?

r = radius of circle = length of rope = 1.75 m

ω = angular speed = (3 rev/s)(2π rad/1 rev) = 18.84 rad/s

Therefore,

v = (1.75 m)(18.84 rad/s)

v = 32.98 m/s

Now, we apply the 3rd equation of motion on the ball, when it breaks:

2gh = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign due to upward motion)

h = height covered = ?

Vf = Final Velocity = 0 m/s (since, the ball finally stops at highest point for a moment)

Vi = Initial Velocity = 32.98 m/s

Therefore,

2(- 9.8 m/s²)h = (0 m/s)² - (32.98 m/s)²

h = ( - 1088.12 m²/s²)/( - 19.6 m/s²)

h = 55.5 m

since, the ball was initially at a height of 2.1 m from ground. So, the total height from ground, will now become:

h = 55.5 m + 2.1 m

<u>h = 57.6 m</u>

7 0
3 years ago
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