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2 years ago

Which equations can be used to solve for acceleration

Physics

2 answers:

dimulka [17.4K]2 years ago

6 0

Answer:

i think its b

Explanation:

Flauer [41]2 years ago

5 0

$a = v \div \: t$
acceleration equals the final velocity divided by time. have a good day!

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Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

2 years ago

what is the force per meter on a straight wire carrying 5.0 a when it is placed in a magnetic field of 0.020 t

Dvinal [7]

The Force per meter on a straight wire carrying current in a magnetic field is 0.045 N/m.

Calculation:-

F/ℓ = B I sin θ

Where B – Magnetic field = 0.02 T I – Current = 5 A

Substituting the values

F/ℓ = (0.02) (5) (sin 27 deg)

F/ℓ = 0.045 N/m

A force is an influence that can alternate the motion of an item. A force can cause an item with mass to trade its pace, i.e., to boost up. force can also be described intuitively as a push or a pull. A pressure has both value and course, making it a vector quantity.

The push or pull on an item with mass causes it to change its velocity. force is an external agent capable of converting a frame's nation of relaxation or motion. It has significance and a path. A force is a push or pulls among gadgets. it is called an interplay because if one object acts on some other, its movement is matched with the aid of a reaction from the alternative object.

Learn more about force here:-brainly.com/question/12970081

#SPJ4

3 0

6 months ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N