Answer:
1931.7N
Step-by-step explanation:
We are told in the question that :
An auto ( a car) weighs = 2500 pounds
It s inclined at n horizontal Ange of 10°
We are asked to find the force that would prevents it from rolling down the street.
Since the unit for Force = Newton or kgm/s²
Step 1
Convert Weight in pounds to kg
1 pound = 0.453592kg
2500 pounds =
2500 pounds × 0.453592kg
= 1133.981kg
Step 2
Find the force that would prevents it from rolling down the street.
Force = Mass × Acceleration due to gravity × sin θ
Acceleration due to gravity = 9.81m/s
Force = 1133.981kg × 9.81 × sin 10°
Force = 1931.7237321 N
Approximately = 1931.7N
Answer:
Midpoint of side EF would be (-.5,4.5)
Step-by-step explanation:
We know that the coordinates of a mid-point C(e,f) of a line segment AB with vertices A(a,b) and B(c,d) is given by:
e=a+c/2,f=b+d/2
Here we have to find the mid-point of side EF.
E(-2,3) i.e. (a,b)=(2,3)
and F(1,6) i.e. (c,d)=(1,6)
Hence, the coordinate of midpoint of EF is:
e=-2+1/2, f=3+6/2
e=-1/2, f=9/2
e=.5, f=4.5
SO, the mid-point would be (-0.5,4.5)
Answer:
2 1/2
Step-by-step explanation:
6/4=3/2=1 1/2
Answer:
S(t) = -4.9t^2 + Vot + 282.24
Step-by-step explanation:
Since the rocket is launched from the ground, So = 0 and S(t) = 0
Using s(t)=gt^2+v0t+s0 to get time t
Where g acceleration due to gravity = -4.9m/s^2. and
initial velocity = 39.2 m/a
0 = -4.9t2 + 39.2t
4.9t = 39.2
t = 8s
Substitute t in the model equation
S(t) = -49(8^2) + 3.92(8) + So
Let S(t) =0
0 = - 313.6 + 31.36 + So
So = 282.24m
The equation that can be used to model the height of the rocket after t seconds will be:
S(t) = -4.9t^2 + Vot + 282.24
Answer: 20x
Step-by-step explanation:
First we should note that 60 minutes = 1 hour. Therefore, 15 minutes = 15/60 = 1/4 hour
Since Jean can pack 5 cartons in 1/4 hours, the number of carton that he can pack in one hour will be:
= 5 ÷ 1/4
= 5 × 4
= 20 cartons.
Therefore, to get the number of cartons that Jean can pack in x hours, we will multiply 20 by x. This will be:
= 20 × x
= 20x
