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alisha [4.7K]
3 years ago
8

a ferryboat has an acceleration of 0.50 m/s. If the ferry travels 125m in 20 seconds while accelerating, what was its initial ve

locity
Physics
1 answer:
Papessa [141]3 years ago
5 0

as we know that

acceleration = 0.50 m/s^2

distance traveled = 125 m

time taken = 20 s

now we will use the equation of kinematics

d = v_i*t + \frac{1}{2}at^2

125 = v_i* 20 + \frac{1}{2}*0.50*20^2

125 = 20*v_i + 100

125 - 100 = 20*v_i

v_i = 1.25 m/s

so its initial speed must be 1.25 m/s

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C. Magnetism

It is a physical property, not a chemical property.

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3 years ago
What is capacitance?
Otrada [13]

Answer:

A, the amount of charge stored per volt

Explanation:

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C = Q/V

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5 0
4 years ago
Determine the angle between the directions of vector a = 3.00i 1.00j and vector b = 1.00i 3.00j .
Katena32 [7]

The angle between the two vectors is 126° 52' 11".

The given parameters;

vector A = 3.00i + 1.00j

vector B = 1.00i + 3.00j

The angle between the two vectors is calculated as follows;

cos  \theta = \frac{A.B}{|A|.|B|}

The dot product of vectors A and B is calculated as;

A.B = ( 3i + 1j ) . ( 1i +3j )

      = ( 3 × 1 ) + ( 1  × 3 )

      = 3 + 3

      = 6

The magnitude of vectors A and B is calculated as;

|A|  = \sqrt[]{3^2 + 1^2} = \sqrt[]{10} \\

|B| = \sqrt[]{1^2 + 3^2} = \sqrt[]{10} \\

The angle between in two vectors is calculated as;

Cos \theta = \frac{6}{\sqrt[]{10}\sqrt[]{10}  } \\\\Cos \theta = \frac{6}{\ 10 }\\\\Cos \theta = 0.6\\\\\theta = cos^-^1 (0.6)\\\\\theta = 126\\

Therefore, the angle between the two vectors is 126° 52' 11".

Learn more about vectors here:

brainly.com/question/25705666

#SPJ4

6 0
2 years ago
I’ll name u BRAINLIEST if u get this right
nekit [7.7K]
V = m1 u1 - m2 u2 / m1
v = 0.01 * 500 - 2 * 1.4 / 0.01
v = 220 m/s
7 0
3 years ago
a crowbar having the lenght of 1.75 is used to balance a load of 500N if the distance between the fulcrum and the load is 0.5m f
Lemur [1.5K]

The effort applied = 200 N

<h3>Further explanation</h3>

Equilibrium :

\tt F_1.d_1=F_2.d_2

Total length = 1.75

The distance between the fulcrum and the load is 0.5 m ⇒ d₁=0.5 m

The distance between the fulcrum and the force applied :

\tt 1.75-0.5=1.25~m\rightarrow d_2=1.25~m

A load of 500N ⇒ F₁=500 N

The force applied :

\tt 500\times 0.5=F_2\times 1.25\\\\F_2=\dfrac{500\times 0.5}{1.25}=200~N

7 0
3 years ago
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