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strojnjashka [21]
2 years ago
13

Answer ?

Physics
1 answer:
morpeh [17]2 years ago
5 0

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

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TEA [102]

Answer:

agricultural production of food

Explanation:

5 0
3 years ago
What causes water to move from the liquid part of the hyrdrosphere to the cyrosphere?
bezimeni [28]

Answer:

When the liquid moves through the hydrosphere, the water collects into a cloud. When it falls to the earth, turning into snow and sleet collecting in rivers and lakes.

Explanation:

Hope that helps

7 0
2 years ago
Read 2 more answers
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
3 years ago
A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t
Pepsi [2]

Answer:

W =84.6\ Nm

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

 dW = F dx

\int dW = F\int dx

W = F\int_0^6 dx

W = F[x]_0^6

W = 14.1 \times 6

W =84.6\ Nm

hence, work done by the force is equal to W =84.6\ Nm

7 0
3 years ago
8. If horizontal velocity is 5 m/s, and vertical velocity is 8 m/s, what is the
zhenek [66]

Answer:

9.4

Explanation:

magnitude is the sum of the squares.

\sqrt{5^2+8^2} =9.4339\\

If you are given horizontal and vertical components, treat those as the rise and run of a triangle, the rise of 8 with a run of 5 and you want to find the hypotenuse.

How do you find the long side of a triangle?  

7 0
3 years ago
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