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Marrrta [24]
3 years ago
14

What formula is used to find an objects acceleration

Physics
1 answer:
ololo11 [35]3 years ago
7 0
Acceleration = velocity / time.


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Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that
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Answer:

a=0.6\ m/s^2

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, a=0.8\ m/s^2

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

m=\dfrac{F}{a}

m=\dfrac{100\ N}{0.8\ m/s^2}

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

(F-F')=ma

(100-25)=125\times a

a=\dfrac{75}{125}=0.6\ m/s^2

So, the new acceleration of the block is 0.6\ m/s^2. Hence, this is the required solution.

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Complete the statements about the law of conservation of momentum.
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Answer:

complete the statements

Explanation:

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Ax = 44.4 m and Ay = 25.1 m<br> Find the magnitude of the<br> vector.
max2010maxim [7]

Answer:

The magnitude of the vector A is <u>51 m.</u>

Explanation:

Given:

The horizontal component of a vector A is given as:

A_x=44.4\ m

The vertical component of a vector A is given as:

A_y=25.1\ m

Now, we know that, a vector A can be resolved into two mutually perpendicular components; one along the x axis and the other along the y axis. The magnitude of the vector A can be written as the square root of the sum of the squares of each component.

Therefore, the magnitude of vector A is given as:

|\overrightarrow A|=\sqrt{A_{x}^2+A_{y}^2}

Now, plug in 44.4 for A_x, 25.1 for A_y and solve for the magnitude of A. This gives,

|\overrightarrow A|=\sqrt{(44.4)^2+(25.1)^2}\\|\overrightarrow A|=\sqrt{1971.36+630.01}\\|\overrightarrow A|=\sqrt{2601.37}\\|\overrightarrow A|=51\ m

Therefore, the magnitude of the vector A is 51 m.

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3 years ago
The incoming infrared radiation from the Sun as it travels towards Earth is _____
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