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adell [148]
3 years ago
12

Physics question i appreciate your help please

Physics
1 answer:
den301095 [7]3 years ago
4 0

Solution: From the given question, we shall find the vector quantity among the

(A) Time , (B) Velocity, (C) Distance , (D) Speed

Concept: <u>Vector Quantity: </u>All those physical quantities which have magnitude as well as specific directions, are called Vector Quantities.

Here, Time, Distance and Speed have only magnitude but have no directions so they will be scalar quantities.

Now, <u>Velocity:</u> It is defined as the change in displacement per unit time. Since the change in the displacement will be in particular direction only. Hence, velocity will be the vector quantity.

Hence, the option (B) Velocity will be the correct option.

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5 0
3 years ago
Look at the image to answer the question correctly.
r-ruslan [8.4K]

Answer:

1-  b: 2- a : 3- c : 4- d

Explanation:

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3 0
3 years ago
As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant
Alika [10]

Answer:

Isolated or Closed system, both are correct

4 0
3 years ago
An electric current in a metal consists of moving
Alik [6]
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6 0
3 years ago
A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
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