Answer: 10.29 sec.
Explanation:
Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.
If the car starts from rest, this means the following:
ΔK = 1/2 m*vf ²
As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:
100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec
Now, we calculate the change in energy:
ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J
<h2>If P= ΔK/Δt, </h2><h2>Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.</h2>
Explanation:
If I guessed, I'd say the answer is choice (a).
The change in skills required to perform new tasks in manufacturing, along with import competition and a decline in mobility, have contributed to the decline of employment rate for manufacturing since 2000
Answer:
The correct option is;
Neither A nor B
Explanation:
The location of the where the thread wears in tire that has too high inflation is at the thread pattern center due to the reduced size of the contact patch with the load of the car resting on the central portion of the tire's contact surface
When the wear occurs at the outer edges of the tire, the load of the car rests on the outer edges as the contact patch increases due to the tire being under-inflated
Camber is the slope provided in road pavement to drain off water from the road
Roads with camber has a raised middle portion and wear due to camber includes outer-edge tread wear, inner-edge tread wear and tire feathering
Answer:
See explaination
Explanation:
int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }
