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olga2289 [7]
3 years ago
10

The sides of a rectangle are 6.01 meters and 12 meters. Taking the significant figures into account, what is the area of the rec

tangle?
OA. 70 square meters

OB. 72 square meters

OC. 72.00 square meters

OD. 72.1 square meters

OE. 72.12square meters
Physics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer: The area of the rectangle is 72.12 square meters

Explanation:

The area of the rectangle is  length x width  

Area = 6.01 m x 12 m = 72.12 m²

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A train accelerates from 23m/s to 190m/s in 54 seconds. What was its acceleration?
Ierofanga [76]

Answer:

Use the method on the image and solve it.

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4 years ago
Please help me with questions 1, 2 and 3. <br> i need a step by step explanation
kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) x=\frac{1}{2}at^2+v_0t+x_0

The equation for velocity v and a constant acceleration a is:

(ii) v=at+v_0

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) a = \frac{v -v_0}{t}

(iv) x = \frac{v-v_0}{2t}t^2+v_0t + x_0

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}

5 0
3 years ago
Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200
artcher [175]
We have all the charges for q1, q2, and q3. 
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 |   / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2 
r^2 x F(3on1) = k |q1 q3| 
r = sqrt of k |q1 q3| / F(3on1) 
= .144 m (distance between q1 and q3)
0 - .144m 

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
6 0
3 years ago
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mr_godi [17]

Answer:

5 Days to Seconds = 432000

Explanation:

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3 years ago
Is it possible to put a distance versus time graph to be universal go mine
xz_007 [3.2K]
Yes I'm pretty sure you can
3 0
3 years ago
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