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3 years ago

How to find acceleration?

2 answers:

4 0

Acceleration = Change in Velocity / time

a = (v - u) / t

Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²

A proper record of the changes in velocity with the corresponding time would help find the acceleration.

AlladinOne [14]3 years ago

4 0

Answer:

Acceleration is given by the formula :

a = (v - u)/t

a is the acceleration , u is the initial velocity , v is the final velocity , t is the time in seconds

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If a Ball Falls from a girls hand, How does its speed change?

Lapatulllka [165]

Answer:

When an object is in free fall, gravity increases its velocity by 9.8 m/s with every passing second.

Explanation:

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7 0

3 years ago

Read 2 more answers

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa

Burka [1]

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

coefficient of static friction μ = 0.40

coefficient of kinetic friction = 0.20

a). when no horizontal force is applied .

according to Newton 's third law of motion If there is no force applied to the box,so the frictional force exerted is 0 N

b) magnitude of friction force

box and the box is initially at rest

friction force =.Static friction coefficient × weight of the box

friction force = 0.40 × 6

friction force = 2.4 N

8 0

3 years ago

Yall have helped a lot i just need help on this then ill be done for a while

Lubov Fominskaja [6]

Answer:

f(x)=a(x - h)2 + k

Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.

The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):

x - h = 0

-2 - h = 0

-h = 2

h = -2

So the function ends up looking like:

f(x)=a(x - (-2))2 + 2

Subtracting a negative cancels the signs out to make a positive:

f(x)=a(x + 2)2 + 2Explanation:

6 0

3 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

Because CMB radiation was _____ throughout the universe, scientists can only relate the source to the big bang.

erma4kov [3.2K]

Answer: 1 The correct answer is that CMB radiation was spread uniformly throughout the whole universe.This was related to big band theory because this theory predicts that the universe was a very hot place and as it cooled down it should have been filled with laterally the remnant heat over from the Big Bang called as cosmic microwave background.

Answer: 2 CMB radiation was discovered accidentally when Penzias and Wilson were performing some experiment and they noticed a ' hum' noise that was constantly detected by the antenna even after removing all the disturbing sources.

Then it was realized that it is cosmic microwave background radiation.

6 0

3 years ago