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viktelen [127]
3 years ago
14

4. What are three tips parents can use to improve communication with their children? Discuss one that you think is most importan

t
Physics
1 answer:
Mumz [18]3 years ago
4 0
Have them give up there phone for 15 minutes
talk to them when there in a good mood
watch a movie then have a disscution about it



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How long do elephants live?
ANTONII [103]
65 years but anything can happen to them
I’m not really sure but I hope this helps
5 0
3 years ago
Read 2 more answers
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in
insens350 [35]

Answer:

659.01W

Explanation:

The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be

Wc = mass of the cab × acceleration due to gravity in m/s²

Wc = 1250 × 9.81 = 12262.5 N

but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N

Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N

the motor of the elevator will have to provide this in form of work

work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m

power provided by the motor of the elevator = workdone by the motor / time in seconds

Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W

5 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

8 0
3 years ago
A circuit has a current of 3.6 A and a resistance of 5.0 Ω.
Sergio039 [100]
If a circuit has a current of 3.6 Amps and resistance of 5 Ohms, then Ohm's law can be used to find the voltage. Ohm's law states that the voltage is equal to the product of current and resistance (V=IR). In this case the voltage is equal to 3.6 Amps x 5 Ohms = 18.0 Volts. The law can also be used with the rearranged equation to obtain current or resistance. 
3 0
3 years ago
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