Answer:
A×B=C×D
500×0.5=250×X
250=250×X
X=250/250=1
X=1 m
Explanation:
note: if the force plus two, the distance will be half.
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s
Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
Answer: 1,600 seconds
Explanation:
31,360/9.8 = 3,200.
Then divide 3,200/2 = 1,600
Answer:
Explanation:
Since,
<h3><u>1 kWh = 1 unit</u></h3>
So,
1.6 kWh = 1.6 units
If,
<h3>1 unit = 9p</h3>
1.6 units = 9p × 1.6
1.6 units = 14.4p
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