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2 years ago

Please help correct answering get 20 points (physical science btw )

Physics

1 answer:

Zinaida [17]2 years ago

6 0

Answer:

Acceleration = 3.67 m/s²

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Given the following data;

Initial velocity, u = 5 m/s

Final velocity, v = 27

Time, t = 6

To find the acceleration;

Acceleration = (v - u)/t

Acceleration = (27 - 5)/6

Acceleration = 22/6

Acceleration = 3.67 m/s²

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3 years ago

Calculate the Force required to give a bullet of mass 50 g an acceleration of 300 m/s2

NNADVOKAT [17]

Answer:

<h3>The answer is 15 N</h3>

Explanation:

The force acting on an object can be found by using the formula

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From the question

mass = 50 g = 0.05 kg

acceleration = 300 m/s²

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force = 0.05 × 300

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3 years ago

Thomas needs to move an 80 kg rock, but cannot lift it. He decides to use a

ArbitrLikvidat [17]

Answer:

Explanation:

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3 years ago

A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude

irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b. $\mid A\mid=46 m$

$\theta=20^{\circ}$ below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis= $x=360-20=340^{\circ}$

x-component of vector A= $A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24$

y-Component of vector A= $A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64$

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis= $x'=42^{\circ}$

x-component of vector B= $B_x=86cos42=63.64$

y-Component of vector B= $B_y=86sin42=57.62$

Vector A= $A_xi+A_yj=43.24i-15.64j$

Vector B= $B_xi+B_yj=63.64i+57.62j$

Vector C=A+B

Substitute the values

$C=43.24i-15.64j+63.64i+57.62j$

$C=106.88i+41.98j$

c.Direction= $\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}$

The direction of the vector C=21.5 degree

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3 years ago

The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin

alekssr [168]

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = $200\mu m = 200\times 10^{- 6}\ m$

Gauge Pressure inside, $P_{in} = 25\ mmHg$

Blood Pressure outside, $P_{o} = 10\ mmHg$

Now,

Change in pressure, $\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa$

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

$\Delta P = \frac{4\pi T}{R}$

$T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m$

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

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