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Nutka1998 [239]
3 years ago
15

A car traveling at 30 m/s north slows to a stop in 15 seconds. What is the acceleration? PLEASE HELP

Physics
2 answers:
Lapatulllka [165]3 years ago
6 0
The answer is C..........
Scilla [17]3 years ago
3 0
The "acceleration" is negative so it's a deceleration. Change of speed = 30 m/s. Time taken for change 15 seconds Deceleration 30m/s^-1 divided by 15 seconds which is 2m/s per second ...
Nice and gentle, and so not the "modern way"
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A car travels 240km in 4h what’s the cars velocity
Tresset [83]

The car's speed is 240km/4hr= 60km/hr.

There's not enough information given in the question to determine its velocity.

6 0
3 years ago
Read 2 more answers
If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
What is 18 m/s north an example of?
Mkey [24]

Answer:

It is an example of velocity

Explanation:

It is an example of velocity Don't ask how I know because I do know it I just don't know how to explain it.

8 0
3 years ago
A
Natasha_Volkova [10]

Answer:

i think it might be D or C

4 0
3 years ago
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