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2 years ago

Any change in an objects speed and /or direction ?

Physics

1 answer:

Alexeev081 [22]2 years ago

8 0

Answer:

The answer to this question is acceleration or as scientist say velocity.

Explanation: We know that when you drive a car the acceleration is effected by the amount of speed you add when you press the gas,so you are accelerating the car faster.

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lood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. As a hemoglobi

uysha [10]

Answer:

$P_1 - P_2 =219.62\ Pa$

Explanation:

given,

density of blood = 1060 kg/m³

v₂ = 0.800 m/s

v₁ = 0.475 m/s

change in pressure calculation

using Bernoulli's equation

$P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$

both are at same level

$P_1 - P_2 =\dfrac{1}{2}\rho (v_2^2- v_1^2)$

$P_1 - P_2 =\dfrac{1}{2}1060\times (0.8^2- 0.475^2)$

$P_1 - P_2 =219.62\ Pa$

the change in pressure is equal to 219.62 Pa

5 0

3 years ago

Define statistics and give an example of three types of variables that researchers study using statistics.

MA_775_DIABLO [31]

This is the answer you’re looking for

8 0

2 years ago

A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to

kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 = 192 + 56 d + 4d^2 - 192

93 = 56 d + 4d^2

4d^2 + 56 d - 93 = 0

$d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}$

$d=\frac{-56\pm 87.72}{8}$ \

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0

3 years ago

Drag each label to the correct location on the chart.

Soloha48 [4]

Kinetic energy-flashlight, and guitar
potential-coal,and composed spring

7 0

3 years ago

Read 2 more answers

A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

3 years ago