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9966 [12]
3 years ago
14

Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is

cm/s. The speed of the cart after 3 seconds of Medium fan speed is cm/s. The speed of the cart after 6 seconds of High fan speed is cm/s.
Physics
2 answers:
Colt1911 [192]3 years ago
8 2

Answer:

The speed of the cart after 3 seconds of Low fan speed is 54.0 cm/s

The speed of the cart after 5 seconds of Medium fan speed is 120.0 cm/s

The speed of the cart after 2 seconds of High fan speed is 64.0 cm/s

Explanation:

just did it on edge

finlep [7]3 years ago
4 0

Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

The speed of the cart after 3 seconds of Medium fan speed is   36.0 cm/s

The speed of the cart after 6 seconds of High fan speed is  96.0 cm/s

Explanation:

took the test on edgenuity

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You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2​
kumpel [21]

Answer:

6

Explanation:

120/30=6

8 0
2 years ago
Mathphys :( im sorry i annoy you
Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

3 0
3 years ago
A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
Please help i'm going to throw up from stress
Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})} where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)} and

16.0=\frac{83640x+2928}{4182x+292.8} and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0
3 years ago
You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas
sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 \frac{a}{g}=\frac{25}{196}

a=1.2512m/sec^2

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m

7 0
4 years ago
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