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True [87]
3 years ago
6

Using a diagram explain how acceleration can be obtained from a velocity time graph​

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

For any v-t graph the acceleration is the slope of the graph. For average acceleration in a time period ‘t’ consider the change in velocity in time t and divide it by the time t. For instantaneous acceleration you need to go into the realm of differential calculus.

Explanation:

I hope this helps! I would really appreciate it if you would please mark me brainliest! Have a blessed day!

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Can an objects displacement be greater than or equal to the objects distance?
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If you say displacement is greater than distance, you will contradict the above statement. Displacement is always less than or equal to distance. Note that distance is a scalar whereas displacement is a vector.So' displacement cannot be more than distance.
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3 years ago
A water strider bug is supported on the surface of a pond by surface tension acting along the interface between the water and th
Alik [6]

Answer:

minimum interface length = 1.36 mm

Explanation:

given data

weight of the bug = 10^{-4} N

solution

we will apply here Surface Tension formula that is

Surface Tension ,σ = Force ÷ length      ........................1

and we consider here surface tension for water is 7.34 × 10^{-2} N/m

so that here minimum interface length needed to support the bug is

minimum interface length = Force ÷ σ

minimum interface length = 10^{-4}  ÷ 7.34 × 10^{-2}

minimum interface length = 1.36 mm

                                                                                     

                                                                                     

4 0
3 years ago
A girl weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm un
8_murik_8 [283]

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

3 0
3 years ago
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