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CaHeK987 [17]
3 years ago
13

If you wanna make it easier you can number them going down 1,2,3, or 4 & just tell me the number

Physics
1 answer:
prisoha [69]3 years ago
8 0

the number is 1 I hope this help

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What happens if magnesium is added to hydrochloric acid (hci) in a single replacement reaction
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It produces hydrogen gas
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2. A hanging wind-chime on a calm day would have kinetic or potential energy?
rjkz [21]

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it would have potential energy

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What a measurement? Write down t its importance inn an our daily life.​
balu736 [363]

Answer:

Time, size, distance, speed, direction, weight, volume, temperature, pressure, force, sound, light, energy—these are among the physical properties for which humans have developed accurate measures, without which we could not live our normal daily lives. Measurement permeates every aspect of human life.

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3 years ago
A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and
iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g

The speed at this point would then be

v^2 = a_c r = 2gr = 2*9.8*17 = 333.2

v = \sqrt{333.2} = 18.25 m/s

2. Similarly, if T = mg, then the centripetal acceleration must be

a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0

As the ball has no centripetal acceleration, its speed must also be 0 as well.

6 0
4 years ago
Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ
Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

v=v_o+at    (1)

x=v_ot+\frac{1}{2}at^2   (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s

Next, you solve the equation (1) for the acceleration a:

a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0
3 years ago
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