How man significant figures does the following value gave 43.023

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0

2 years ago

FIRST ANSWER IS BRAINIEST TAKE YOUR TIME I NEED RIGHT ANSWERS FIRST ANSWER IS BRAINIEST ANSWER TAKE YOUR TIME NEED RIGHT ANSWER

V125BC [204]

A = 7
B = 4
C = 5
D = 9
E = 8
F = 1
G = 2
H = 10
I = 6
J = 3

4 0

2 years ago

Read 2 more answers

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

4 0

2 years ago

WOULUJUTUL RECIPECUIUS.

3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider $M_{1}$ and $M_{2}$ as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

$\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}$

As gravitational constant $G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}$ , $M_{1}$ = 20 kg and $M_{2}$ = 100 kg, while d = 2.6 m, then

$\text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}$

Thus, we get finally,

$\text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}$

As we know, nano denoted by letter 'n' equals to $10^{-9}$

So the force acting between two objects is 19.73 nN.

7 0

2 years ago

Which type of telescope does not require darkness in order to be able to use it?

exis [7]

Answer:

A type of telescope that does not require darkness in order to be able to use it is the refracting telescope

Explanation:

A refracting telescope consists of a lens and an eyepiece collects light which is then focused to present a magnified, bright and clear image.

The incident light on a refracting telescope is bent by refraction such that the light is focused to the focal point.

In refracting telescopes, the image is formed by bending light, that is by refraction.

The refracting telescope technology has been applied to binoculars and camera zoom lenses.

5 0

2 years ago